2025-11-10
Math Analysis Homework - Week 8
2025-11-10 _posts/homework/sem1
Math Analysis Homework - Week 8
Class 1
T1
∫ x + 1 x 3 + 2 x 2 − x − 2 d x \begin{gathered}
\int \dfrac{x+1}{x^3+2x^2-x-2} dx
\end{gathered} ∫ x 3 + 2 x 2 − x − 2 x + 1 d x
= ∫ x + 1 ( x + 1 ) ( x − 1 ) ( x + 2 ) d x = ∫ 1 3 ( 1 x − 1 − 1 x + 2 ) d x = 1 3 ln ∣ x − 1 x + 2 ∣ + C \begin{gathered}
=\int \dfrac{x+1}{(x+1)(x-1)(x+2)}dx \\
=\int \dfrac{1}{3} (\dfrac{1}{x-1} -\dfrac{1}{x+2}) dx \\
=\dfrac{1}{3} \ln \vert \dfrac{x-1}{x+2} \vert +C
\end{gathered} = ∫ ( x + 1 ) ( x − 1 ) ( x + 2 ) x + 1 d x = ∫ 3 1 ( x − 1 1 − x + 2 1 ) d x = 3 1 ln ∣ x + 2 x − 1 ∣ + C
T2
∫ x − 1 ( x 2 + 2 x + 3 ) 2 d x \begin{gathered}
\int \dfrac{x-1}{(x^2+2x+3)^2} dx
\end{gathered} ∫ ( x 2 + 2 x + 3 ) 2 x − 1 d x
let u = x 2 + 2 x + 3 a n s = 1 2 ∫ d u u 2 − 2 ∫ d x ( x 2 + 2 x + 3 ) 2 = − 1 2 ( x 2 + 2 x + 3 ) − 2 ∫ d x ( ( x + 1 ) 2 + 2 2 ) 2 let x + 1 = 2 tan v a n s = − 1 2 ( x 2 + 2 x + 3 ) − 2 ∫ 2 sec 2 v 4 sec 4 v d v = − 1 2 ( x 2 + 2 x + 3 ) − 2 2 ∫ 1 + cos 2 v 2 d v = − 1 2 ( x 2 + 2 x + 3 ) − 2 4 arctan x + 1 2 − 1 2 x + 1 x 2 + 2 x + 3 + C \begin{gathered}
\text{let } u=x^2+2x+3 \\
ans=\dfrac{1}{2} \int \dfrac{du}{u^2} -2\int \dfrac{dx}{(x^2+2x+3)^2} \\
=-\dfrac{1}{2(x^2+2x+3)} -2\int \dfrac{dx}{((x+1)^2+{\sqrt 2}^2)^2} \\
\text{let } x+1=\sqrt 2\tan v \\
ans=-\dfrac{1}{2(x^2+2x+3)} -2\int \dfrac{\sqrt 2 \sec^2 v}{4\sec^4 v}dv \\
=-\dfrac{1}{2(x^2+2x+3)} -\dfrac{\sqrt 2}{2}\int \dfrac{1+\cos 2v}{2} dv \\
=-\dfrac{1}{2(x^2+2x+3)} -\dfrac{\sqrt 2}{4} \arctan\dfrac{x+1}{\sqrt 2} -\dfrac{1}{2}\dfrac{x+1}{x^2+2x+3} +C
\end{gathered} let u = x 2 + 2 x + 3 an s = 2 1 ∫ u 2 d u − 2 ∫ ( x 2 + 2 x + 3 ) 2 d x = − 2 ( x 2 + 2 x + 3 ) 1 − 2 ∫ (( x + 1 ) 2 + 2 2 ) 2 d x let x + 1 = 2 tan v an s = − 2 ( x 2 + 2 x + 3 ) 1 − 2 ∫ 4 sec 4 v 2 sec 2 v d v = − 2 ( x 2 + 2 x + 3 ) 1 − 2 2 ∫ 2 1 + cos 2 v d v = − 2 ( x 2 + 2 x + 3 ) 1 − 4 2 arctan 2 x + 1 − 2 1 x 2 + 2 x + 3 x + 1 + C
T3
∫ x 4 x 4 + 5 x 2 + 4 d x \begin{gathered}
\int \dfrac{x^4}{x^4+5x^2+4} dx
\end{gathered} ∫ x 4 + 5 x 2 + 4 x 4 d x
= ∫ ( 1 − 5 x 2 + 4 ( x 2 + 4 ) ( x 2 + 1 ) ) d x = ∫ ( 1 − ( 16 3 ( x 2 + 4 ) − 1 3 ( x 2 + 1 ) ) ) d x = x − 8 3 arctan x 2 − 1 3 arctan x + C \begin{gathered}
=\int (1-\dfrac{5x^2+4}{(x^2+4)(x^2+1)} )dx \\
=\int (1-(\dfrac{16}{3(x^2+4)}-\dfrac{1}{3(x^2+1)} ))dx \\
=x-\dfrac{8}{3} \arctan\dfrac{x}{2} -\dfrac{1}{3} \arctan x+C
\end{gathered} = ∫ ( 1 − ( x 2 + 4 ) ( x 2 + 1 ) 5 x 2 + 4 ) d x = ∫ ( 1 − ( 3 ( x 2 + 4 ) 16 − 3 ( x 2 + 1 ) 1 )) d x = x − 3 8 arctan 2 x − 3 1 arctan x + C
T4
∫ 1 sin x + tan x d x \begin{gathered}
\int \dfrac{1}{\sin x+\tan x} dx
\end{gathered} ∫ sin x + tan x 1 d x
= 1 4 ∫ ( 1 + tan 2 ( x 2 ) ) ( 1 − tan 2 ( x 2 ) ) tan x 2 d x = 1 4 ∫ ( cot x 2 − tan 3 x 2 ) d x ∫ tan 3 x d x = ∫ sec 2 x tan x d x − ∫ tan x d x = tan 2 x 2 − ln sec x + C A n s = 1 2 ln sin x 2 − tan 2 x 2 4 + 1 2 ln sec x 2 + C \begin{gathered}
=\dfrac{1}{4} \int \dfrac{(1+\tan^2(\dfrac{x}{2}))(1-\tan^2(\dfrac{x}{2}))}{\tan \dfrac{x}{2} } dx \\
=\dfrac{1}{4} \int (\cot \dfrac{x}{2} -\tan^3\dfrac{x}{2} )dx \\
\int \tan^3 xdx \\
=\int \sec^2 x\tan xdx-\int \tan xdx \\
=\dfrac{\tan^2 x}{2} -\ln \sec x+C \\
Ans=\dfrac{1}{2} \ln \sin \dfrac{x}{2} -\dfrac{\tan^2 \dfrac{x}{2} }{4} +\dfrac{1}{2} \ln \sec \dfrac{x}{2} +C
\end{gathered} = 4 1 ∫ tan 2 x ( 1 + tan 2 ( 2 x )) ( 1 − tan 2 ( 2 x )) d x = 4 1 ∫ ( cot 2 x − tan 3 2 x ) d x ∫ tan 3 x d x = ∫ sec 2 x tan x d x − ∫ tan x d x = 2 tan 2 x − ln sec x + C A n s = 2 1 ln sin 2 x − 4 tan 2 2 x + 2 1 ln sec 2 x + C
T5
∫ sin 2 x sin 4 x + cos 4 x d x \begin{gathered}
\int \dfrac{\sin 2x}{\sin^4 x+\cos^4 x} dx
\end{gathered} ∫ sin 4 x + cos 4 x sin 2 x d x
= ∫ sin 2 x ( sin 2 x + cos 2 x ) 2 − 1 2 sin 2 2 x d x = ∫ sin 2 x 1 2 + 1 2 cos 2 2 x d x = − ∫ d cos 2 x 1 + cos 2 2 x d x = − arctan cos 2 x + C \begin{gathered}
=\int \dfrac{\sin 2x}{(\sin^2 x+\cos^2 x)^2-\dfrac{1}{2} \sin^2 2x} dx \\
=\int \dfrac{\sin 2x}{\dfrac{1}{2} +\dfrac{1}{2} \cos^2 2x} dx \\
=-\int \dfrac{d\cos 2x}{1+\cos^2 2x} dx \\
=-\arctan \cos 2x+C
\end{gathered} = ∫ ( sin 2 x + cos 2 x ) 2 − 2 1 sin 2 2 x sin 2 x d x = ∫ 2 1 + 2 1 cos 2 2 x sin 2 x d x = − ∫ 1 + cos 2 2 x d cos 2 x d x = − arctan cos 2 x + C
T6
∫ 1 1 + 1 + x 3 d x \begin{gathered}
\int \dfrac{1}{1+\sqrt[3]{1+x}} dx
\end{gathered} ∫ 1 + 3 1 + x 1 d x
let t = 1 + x 3 , x = t 3 − 1 a n s = ∫ 3 t 2 d t t + 1 = 3 ∫ ( t − 1 + 1 t + 1 ) d t = 3 ( t 2 2 − t + ln ( t + 1 ) ) + C = 3 2 ( 1 + x ) 2 3 − 3 ( 1 + x ) 1 3 + 3 ln ( 1 + 1 + x 3 ) + C \begin{gathered}
\text{let } t=\sqrt[ 3 ]{ 1+x },x=t^3-1 \\
ans=\int \dfrac{3t^2dt}{t+1} \\
=3\int (t-1+\dfrac{1}{t+1}) dt \\
=3(\dfrac{t^2}{2}-t+\ln(t+1))+C \\
=\dfrac{3}{2} (1+x)^{\frac23}-3(1+x)^{\frac13}+3\ln(1+\sqrt[ 3 ]{ 1+x } )+C
\end{gathered} let t = 3 1 + x , x = t 3 − 1 an s = ∫ t + 1 3 t 2 d t = 3 ∫ ( t − 1 + t + 1 1 ) d t = 3 ( 2 t 2 − t + ln ( t + 1 )) + C = 2 3 ( 1 + x ) 3 2 − 3 ( 1 + x ) 3 1 + 3 ln ( 1 + 3 1 + x ) + C
T7
∫ 1 x 1 + x 2 d x \begin{gathered}
\int \dfrac{1}{x\sqrt{1+x^2}} dx
\end{gathered} ∫ x 1 + x 2 1 d x
let x = tan t a n s = ∫ sec 2 t tan t sec t d t = ∫ csc t d t = ln ∣ csc t − cot t ∣ + C = ln ∣ x 2 + 1 − 1 x ∣ + C \begin{gathered}
\text{let } x=\tan t \\
ans=\int \dfrac{\sec^2 t}{\tan t \sec t} dt \\
=\int \csc t dt \\
=\ln \vert \csc t-\cot t \vert +C \\
=\ln \vert \dfrac{\sqrt{ x^2+1 } -1}{x} \vert +C
\end{gathered} let x = tan t an s = ∫ tan t sec t sec 2 t d t = ∫ csc t d t = ln ∣ csc t − cot t ∣ + C = ln ∣ x x 2 + 1 − 1 ∣ + C
T8
∫ 1 x + x 2 + x + 1 d x \begin{gathered}
\int \dfrac{1}{x+\sqrt{x^2+x+1}} dx
\end{gathered} ∫ x + x 2 + x + 1 1 d x
let x 2 + x + 1 = x + t x 2 + x + 1 = x 2 + 2 x t + t 2 x = t 2 − 1 1 − 2 t ⟹ a n s = ∫ 1 2 t 2 − 1 1 − 2 t + t 2 t ( 1 − 2 t ) + 2 ( t 2 − 1 ) ( 1 − 2 t ) 2 d t = ∫ t 2 − t + 1 ( t − 2 ) ( t − 1 2 ) d t = ∫ ( 1 + 1 2 ( 4 t − 2 − 1 t − 1 2 ) ) d t = t + 2 ln ( t − 2 ) − 1 2 ln ( t − 1 2 ) + C = x 2 + x + 1 − x + 2 ln ( x 2 + x + 1 − x − 2 ) − 1 2 ln ( x 2 + x + 1 − x − 1 2 ) + C \begin{gathered}
\text{let } \sqrt{x^2+x+1}=x+t \\
x^2+x+1=x^2+2xt+t^2 \\
x=\dfrac{t^2-1}{1-2t} \\
\implies ans=\int \dfrac{1}{2\dfrac{t^2-1}{1-2t} +t} \dfrac{2t(1-2t)+2(t^2-1)}{(1-2t)^2} dt \\
=\int \dfrac{t^2-t+1}{(t-2)(t-\dfrac{1}{2} )} dt \\
=\int (1+\dfrac{1}{2} (\dfrac{4}{t-2} -\dfrac{1}{t-\frac12} ))dt \\
=t+2\ln(t-2)-\dfrac{1}{2} \ln (t-\dfrac{1}{2} )+C \\
=\sqrt{x^2+x+1}-x+2\ln(\sqrt{x^2+x+1}-x-2) \\
-\dfrac{1}{2} \ln(\sqrt{x^2+x+1}-x-\dfrac{1}{2} )+C
\end{gathered} let x 2 + x + 1 = x + t x 2 + x + 1 = x 2 + 2 x t + t 2 x = 1 − 2 t t 2 − 1 ⟹ an s = ∫ 2 1 − 2 t t 2 − 1 + t 1 ( 1 − 2 t ) 2 2 t ( 1 − 2 t ) + 2 ( t 2 − 1 ) d t = ∫ ( t − 2 ) ( t − 2 1 ) t 2 − t + 1 d t = ∫ ( 1 + 2 1 ( t − 2 4 − t − 2 1 1 )) d t = t + 2 ln ( t − 2 ) − 2 1 ln ( t − 2 1 ) + C = x 2 + x + 1 − x + 2 ln ( x 2 + x + 1 − x − 2 ) − 2 1 ln ( x 2 + x + 1 − x − 2 1 ) + C
T9
∫ 1 ( 1 + e x ) 2 d x \begin{gathered}
\int \dfrac{1}{(1+e^x)^2}dx
\end{gathered} ∫ ( 1 + e x ) 2 1 d x
let t = e x = ∫ 1 t ( 1 + t ) 2 d t = ∫ ( 1 t − 1 ( 1 + t ) 2 − 1 1 + t ) d x = ln t t + 1 + 1 1 + t + C = ln e x 1 + e x + 1 1 + e x + C \begin{gathered}
\text{let }t=e^x \\
=\int \dfrac{1}{t(1+t)^2}dt \\
=\int (\dfrac{1}{t} -\dfrac{1}{(1+t)^2} -\dfrac{1}{1+t}) dx \\
=\ln \dfrac{t}{t+1} +\dfrac{1}{1+t} +C \\
=\ln \dfrac{e^x}{1+e^x} +\dfrac{1}{1+e^x} +C
\end{gathered} let t = e x = ∫ t ( 1 + t ) 2 1 d t = ∫ ( t 1 − ( 1 + t ) 2 1 − 1 + t 1 ) d x = ln t + 1 t + 1 + t 1 + C = ln 1 + e x e x + 1 + e x 1 + C
T10
∫ x ln 1 + x 1 − x d x \begin{gathered}
\int x\ln\dfrac{1+x}{1-x} dx
\end{gathered} ∫ x ln 1 − x 1 + x d x
= ∫ ( ( 1 + x ) ln ( 1 + x ) − ln ( 1 + x ) + ( 1 − x ) ln ( 1 − x ) − ln ( 1 − x ) ) d x ∫ x ln x d x = x 2 ln x 2 − x 2 4 + C ∫ ln x d x = x ln x − x + C ⟹ a n s = ( 1 + x ) 2 4 ( 2 ln ( 1 + x ) − 1 ) − ( 1 + x ) ln ( 1 + x ) + ( 1 + x ) − ( 1 − x ) 2 4 ( 2 ln ( 1 − x ) − 1 ) + ( 1 − x ) ln ( 1 − x ) − ( 1 − x ) + C = x 2 − 1 2 ln 1 + x 1 − x + x + C \begin{gathered}
=\int ((1+x)\ln(1+x)-\ln(1+x)+(1-x)\ln(1-x)-\ln(1-x))dx \\
\int x\ln xdx=\dfrac{x^2\ln x}{2} -\dfrac{x^2}{4} +C \\
\int \ln xdx=x\ln x-x+C \\
\implies ans=\dfrac{(1+x)^2}{4} (2\ln (1+x)-1)-(1+x)\ln(1+x)+(1+x) \\
-\dfrac{(1-x)^2}{4} (2\ln(1-x)-1)+(1-x)\ln(1-x)-(1-x) +C\\
=\dfrac{x^2-1}{2} \ln\dfrac{1+x}{1-x} +x+C
\end{gathered} = ∫ (( 1 + x ) ln ( 1 + x ) − ln ( 1 + x ) + ( 1 − x ) ln ( 1 − x ) − ln ( 1 − x )) d x ∫ x ln x d x = 2 x 2 ln x − 4 x 2 + C ∫ ln x d x = x ln x − x + C ⟹ an s = 4 ( 1 + x ) 2 ( 2 ln ( 1 + x ) − 1 ) − ( 1 + x ) ln ( 1 + x ) + ( 1 + x ) − 4 ( 1 − x ) 2 ( 2 ln ( 1 − x ) − 1 ) + ( 1 − x ) ln ( 1 − x ) − ( 1 − x ) + C = 2 x 2 − 1 ln 1 − x 1 + x + x + C
T11
∫ arctan ( 1 + x ) d x \begin{gathered}
\int \arctan(1+\sqrt x)dx
\end{gathered} ∫ arctan ( 1 + x ) d x
= x arctan ( 1 + x ) − ∫ x 1 2 x ( 1 + ( x + 1 ) 2 ) d x let t = x a n s = x arctan ( 1 + x ) − ∫ t 2 d t 1 + ( 1 + t ) 2 = x arctan ( 1 + x ) − ∫ ( d t − d ( t + 1 ) 2 1 + ( 1 + t ) 2 ) = x arctan ( 1 + x ) − t + ln ( 1 + ( 1 + t ) 2 ) + C = x arctan ( 1 + x ) − x + ln ( 1 + ( 1 + x ) 2 ) + C \begin{gathered}
=x\arctan(1+\sqrt x)-\int x\dfrac{1}{2\sqrt x(1+(\sqrt x+1)^2)} dx \\
\text{let } t=\sqrt x \\
ans=x\arctan(1+\sqrt x)-\int \dfrac{t^2dt}{1+(1+t)^2} \\
=x\arctan(1+\sqrt x)-\int (dt-\dfrac{d(t+1)^2}{1+(1+t)^2} ) \\
=x\arctan(1+\sqrt x)-t+\ln(1+(1+t)^2)+C \\
=x\arctan(1+\sqrt x)-\sqrt x+\ln(1+(1+\sqrt x)^2)+C
\end{gathered} = x arctan ( 1 + x ) − ∫ x 2 x ( 1 + ( x + 1 ) 2 ) 1 d x let t = x an s = x arctan ( 1 + x ) − ∫ 1 + ( 1 + t ) 2 t 2 d t = x arctan ( 1 + x ) − ∫ ( d t − 1 + ( 1 + t ) 2 d ( t + 1 ) 2 ) = x arctan ( 1 + x ) − t + ln ( 1 + ( 1 + t ) 2 ) + C = x arctan ( 1 + x ) − x + ln ( 1 + ( 1 + x ) 2 ) + C
12
∫ x sec 2 x ( 1 + tan x ) 2 d x \begin{gathered}
\int \dfrac{x\sec^2 x}{(1+\tan x)^2} dx
\end{gathered} ∫ ( 1 + tan x ) 2 x sec 2 x d x
∫ sec 2 x ( 1 + tan x ) 2 d x = ∫ d tan x ( 1 + tan x ) 2 − 1 1 + tan x + C a n s = − x 1 + tan x + ∫ d x 1 + tan x ∫ d x 1 + tan x = ∫ cos x cos x + sin x d x = 1 2 ∫ ( cos x + sin x ) d x + d ( cos x + sin x ) cos x + sin x = 1 2 ln ( cos x + sin x ) + 1 2 x + C ⟹ a n s = − x 1 + tan x + 1 2 ln ( cos x + sin x ) + 1 2 x + C \begin{gathered}
\int \dfrac{\sec^2x}{(1+\tan x)^2} dx \\
=\int \dfrac{d\tan x}{(1+\tan x)^2} \\
-\dfrac{1}{1+\tan x} +C \\
ans=-\dfrac{x}{1+\tan x} +\int \dfrac{dx}{1+\tan x} \\
\int \dfrac{dx}{1+\tan x} \\
=\int \dfrac{\cos x}{\cos x+\sin x} dx \\
=\dfrac{1}{2} \int \dfrac{(\cos x+\sin x)dx+d(\cos x+\sin x)}{\cos x+\sin x} \\
=\dfrac{1}{2} \ln(\cos x+\sin x)+\dfrac{1}{2} x+C \\
\implies ans=-\dfrac{x}{1+\tan x} +\dfrac{1}{2} \ln(\cos x+\sin x)+\dfrac{1}{2} x+C
\end{gathered} ∫ ( 1 + tan x ) 2 sec 2 x d x = ∫ ( 1 + tan x ) 2 d tan x − 1 + tan x 1 + C an s = − 1 + tan x x + ∫ 1 + tan x d x ∫ 1 + tan x d x = ∫ cos x + sin x cos x d x = 2 1 ∫ cos x + sin x ( cos x + sin x ) d x + d ( cos x + sin x ) = 2 1 ln ( cos x + sin x ) + 2 1 x + C ⟹ an s = − 1 + tan x x + 2 1 ln ( cos x + sin x ) + 2 1 x + C
T13
∫ x 2 1 + x 2 arctan x d x \begin{gathered}
\int \dfrac{x^2}{1+x^2} \arctan xdx
\end{gathered} ∫ 1 + x 2 x 2 arctan x d x
let x = tan t a n s = ∫ t tan 2 t d t ∫ tan 2 d t = ∫ ( sec 2 t − 1 ) d t = tan t − t + C a n s = t ( tan t − t ) − ∫ ( tan t − t ) d t = t tan t − ln sec t − t 2 2 + C = x arctan x − ln t 2 + 1 − 1 2 arctan 2 ( t ) + C \begin{gathered}
\text{let }x=\tan t \\
ans=\int t\tan^2 tdt \\
\int \tan^2 dt=\int (\sec^2 t-1)dt \\
=\tan t-t+C \\
ans=t(\tan t-t)-\int (\tan t-t )dt \\
=t\tan t-\ln \sec t-\dfrac{t^2}{2} +C \\
=x\arctan x-\ln \sqrt{t^2+1}-\dfrac{1}{2} \arctan^2(t)+C
\end{gathered} let x = tan t an s = ∫ t tan 2 t d t ∫ tan 2 d t = ∫ ( sec 2 t − 1 ) d t = tan t − t + C an s = t ( tan t − t ) − ∫ ( tan t − t ) d t = t tan t − ln sec t − 2 t 2 + C = x arctan x − ln t 2 + 1 − 2 1 arctan 2 ( t ) + C
T14
∫ x + sin x 1 + cos x d x \begin{gathered}
\int \dfrac{x+\sin x}{1+\cos x} dx
\end{gathered} ∫ 1 + cos x x + sin x d x
= ∫ ( tan x 2 + x 1 + 1 − tan 2 x 2 1 + tan 2 x 2 ) d x = ∫ tan x 2 + x 2 + tan 2 x 2 x 2 d x let u = x 2 a n s = 2 ∫ ( tan u + u + u tan 2 u ) d u = 2 ( x 2 tan x 2 ) + C = x tan x 2 + C \begin{gathered}
=\int {\left( \tan\dfrac{x}{2} +\dfrac{x}{1+\dfrac{1-\tan^2 \dfrac{x}{2}}{1+\tan^2 \dfrac{x}{2} } } \right)} dx \\
=\int \tan\dfrac{x}{2} +\dfrac{x}{2} +\dfrac{\tan^2 \dfrac{x}{2} x}{2} dx \\
\text{let } u=\dfrac{x}{2} \\
ans=2\int (\tan u+u+u\tan^2 u )du \\
=2 {\left( \dfrac{x}{2} \tan \dfrac{x}{2} \right)} +C \\
=x\tan \dfrac{x}{2} +C
\end{gathered} = ∫ tan 2 x + 1 + 1 + tan 2 2 x 1 − tan 2 2 x x d x = ∫ tan 2 x + 2 x + 2 tan 2 2 x x d x let u = 2 x an s = 2 ∫ ( tan u + u + u tan 2 u ) d u = 2 ( 2 x tan 2 x ) + C = x tan 2 x + C
T15
f ( x 2 − 1 ) = ln x 2 x 2 − 2 , f ( ϕ ( x ) ) = ln x calc ∫ ϕ ( x ) d x \begin{gathered}
f(x^2-1)=\ln \dfrac{x^2}{x^2-2} ,f(\phi(x))=\ln x \\
\text{calc } \int \phi(x)dx
\end{gathered} f ( x 2 − 1 ) = ln x 2 − 2 x 2 , f ( ϕ ( x )) = ln x calc ∫ ϕ ( x ) d x
f ( x ) = ln x + 1 x − 1 , x ≥ − 1 f ( ϕ ( x ) ) = ln x ⟺ ϕ ( x ) + 1 ϕ ( x ) − 1 = x , ϕ ( x ) = x + 1 x − 1 a n s = ∫ x + 1 x − 1 d x = ∫ ( 1 + 2 x − 1 ) d x = x + 2 ln ( x − 1 ) + C \begin{gathered}
f(x)=\ln \dfrac{x+1}{x-1} ,x\ge -1 \\
f(\phi (x))=\ln x \\
\iff \dfrac{\phi(x)+1}{\phi(x)-1} =x,\phi (x)=\dfrac{x+1}{x-1} \\
ans=\int \dfrac{x+1}{x-1} dx \\
=\int (1+\dfrac{2}{x-1}) dx \\
=x+2\ln(x-1)+C
\end{gathered} f ( x ) = ln x − 1 x + 1 , x ≥ − 1 f ( ϕ ( x )) = ln x ⟺ ϕ ( x ) − 1 ϕ ( x ) + 1 = x , ϕ ( x ) = x − 1 x + 1 an s = ∫ x − 1 x + 1 d x = ∫ ( 1 + x − 1 2 ) d x = x + 2 ln ( x − 1 ) + C
Class 2
T1
{ f ( x ) is integrable on [ a , b ] , ∀ [ α , β ] ⊂ [ a , b ] , sup x ∈ [ α , β ] f ( x ) ≥ M ⟹ ∫ a b f ( x ) d x ≥ M ( b − a ) \begin{gathered}
\begin{cases}
f(x) \text{ is integrable on } [a,b], \\
\forall [\alpha,\beta]\subset [a,b],\sup_{x\in [\alpha,\beta]}f(x)\ge M
\end{cases} \\
\implies \int_a^b f(x)dx\ge M(b-a)
\end{gathered} { f ( x ) is integrable on [ a , b ] , ∀ [ α , β ] ⊂ [ a , b ] , sup x ∈ [ α , β ] f ( x ) ≥ M ⟹ ∫ a b f ( x ) d x ≥ M ( b − a )
∀ ϵ > 0 , ∀ T = { a = t 0 < t 1 < t 2 … t n = b } , ∃ ξ i ∈ [ t i − 1 , t i ] s . t . f ( ξ i ) > M − ϵ ⟹ ∑ i = 1 n f ( ξ i ) ( t i − t i − 1 ) ≥ ∑ i = 1 n ( M − ϵ ) ( t i − t i − 1 ) = ( M − ϵ ) ( b − a ) f ( x ) is integrable ⟹ ∫ a b f ( x ) d x = lim ∣ ∣ T ∣ ∣ → 0 ∑ i = 1 n f ( ξ i ) ( t i − t i − 1 ) ≥ ( M − ϵ ) ( b − a ) ⟹ ∫ a b f ( x ) d x = lim ϵ → 0 ∫ a b f ( x ) d x ≥ M ( b − a ) \begin{gathered}
\forall \epsilon>0, \\
\forall T=\{ a=t_0<t_1<t_2\ldots t_n=b \} , \\
\exists \xi_i\in [t_{i-1},t_i] \ s.t.\
f(\xi_i)>M-\epsilon \\
\implies \sum _{i = 1} ^{n} f(\xi_i)(t_i-t_{i-1}) \\
\ge \sum _{i = 1} ^{n} (M-\epsilon)(t_i-t_{i-1}) \\
=(M-\epsilon)(b-a) \\
f(x) \text{ is integrable} \implies \\
\int _a^bf(x)dx \\
=\lim_{\vert\vert T \vert\vert \to 0} \sum _{i = 1} ^{n} f(\xi_i)(t_i-t_{i-1}) \\
\ge (M-\epsilon)(b-a) \\
\implies \int_a^b f(x)dx=\lim_{\epsilon \to 0} \int_a^b f(x)dx\ge M(b-a)
\end{gathered} ∀ ϵ > 0 , ∀ T = { a = t 0 < t 1 < t 2 … t n = b } , ∃ ξ i ∈ [ t i − 1 , t i ] s . t . f ( ξ i ) > M − ϵ ⟹ i = 1 ∑ n f ( ξ i ) ( t i − t i − 1 ) ≥ i = 1 ∑ n ( M − ϵ ) ( t i − t i − 1 ) = ( M − ϵ ) ( b − a ) f ( x ) is integrable ⟹ ∫ a b f ( x ) d x = ∣∣ T ∣∣ → 0 lim i = 1 ∑ n f ( ξ i ) ( t i − t i − 1 ) ≥ ( M − ϵ ) ( b − a ) ⟹ ∫ a b f ( x ) d x = ϵ → 0 lim ∫ a b f ( x ) d x ≥ M ( b − a )
T2
{ f ( x ) = x ( 1 − x ) D ( x ) D ( x ) = [ x ∈ Q ] ⟹ f ( x ) is not integrable on [ 0 , 1 ] \begin{gathered}
\begin{cases}
f(x)=x(1-x)D(x) \\
D(x)=[x\in Q]
\end{cases} \\
\implies f(x) \text{ is not integrable on }[0,1]
\end{gathered} { f ( x ) = x ( 1 − x ) D ( x ) D ( x ) = [ x ∈ Q ] ⟹ f ( x ) is not integrable on [ 0 , 1 ]
T = { a = t 0 < t 1 < … < t n = b } S ( T , ξ ) = ∑ i = 1 n f ( ξ i ) ( t i − t i − 1 ) ∀ i , ∃ ξ i ∈ Q ∩ [ t i , t i − 1 ] , ξ i ′ ∈ Q C ∩ [ t i , t i − 1 ] lim ∣ ∣ T ∣ ∣ → 0 S ( T , ξ ) = ∫ 0 1 x ( 1 − x ) d x ≠ 0 lim ∣ ∣ T ∣ ∣ → 0 S ( T , ξ ′ ) = lim ∣ ∣ T ∣ ∣ → 0 ∑ i = 1 n 0 x ( 1 − x ) ( t i − t i − 1 ) = 0 ≠ lim ∣ ∣ T ∣ ∣ → 0 S ( T , ξ ) ∴ Not integrable \begin{gathered}
T=\{ a=t_0<t_1<\ldots<t_n=b \} \\
S(T,\xi)=\sum _{i = 1} ^{n} f(\xi_i)(t_i-t_{i-1}) \\
\forall i,\exists \xi_i\in Q\cap [t_i,t_{i-1}],\xi_i'\in Q^C\cap [t_i,t_{i-1}] \\
\lim_{\vert\vert T \vert\vert \to 0} S(T,\xi)=\int_0^1 x(1-x)dx\ne 0 \\
\lim_{\vert\vert T \vert\vert \to 0} S(T,\xi')=\lim_{\vert\vert T \vert\vert \to 0}\sum _{i = 1} ^{n} 0x(1-x)(t_i-t_{i-1})=0 \\ \\
\ne \lim_{\vert\vert T \vert\vert \to 0} S(T,\xi) \\
\therefore \text{Not integrable}
\end{gathered} T = { a = t 0 < t 1 < … < t n = b } S ( T , ξ ) = i = 1 ∑ n f ( ξ i ) ( t i − t i − 1 ) ∀ i , ∃ ξ i ∈ Q ∩ [ t i , t i − 1 ] , ξ i ′ ∈ Q C ∩ [ t i , t i − 1 ] ∣∣ T ∣∣ → 0 lim S ( T , ξ ) = ∫ 0 1 x ( 1 − x ) d x = 0 ∣∣ T ∣∣ → 0 lim S ( T , ξ ′ ) = ∣∣ T ∣∣ → 0 lim i = 1 ∑ n 0 x ( 1 − x ) ( t i − t i − 1 ) = 0 = ∣∣ T ∣∣ → 0 lim S ( T , ξ ) ∴ Not integrable