Find the value of int_{0}^{1} frac{log (1+x)} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $I = \int_{0}^{1} \frac{\log (1+x)}{1+x^{2}} d x$.Substitute $x = 	an 	heta$,then $d x = \sec^{2} 	heta d 	heta$.When $x = 0$,$	heta = 0$

Vedclass · Accepted Answer

$\frac{\pi}{8} \log 2$

Vedclass · Answer

(D) Let $I = \int_{0}^{1} \frac{\log (1+x)}{1+x^{2}} d x$.Substitute $x = 	an 	heta$,then $d x = \sec^{2} 	heta d 	heta$.When $x = 0$,$	heta = 0$,and when $x = 1$,$	heta = \frac{\pi}{4}$.Thus,$I = \int_{0}^{\pi/4} \frac{\log (1+	an 	heta)}{1+	an^{2} 	heta} (\sec^{2} 	heta) d 	heta = \int_{0}^{\pi/4} \log (1+	an 	heta) d 	heta \quad ...(i)$Using the property $\int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x$,we have:$I = \int_{0}^{\pi/4} \log \left[1+	an \left(\frac{\pi}{4}-	hetaight)ight] d 	heta$$I = \int_{0}^{\pi/4} \log \left[1+\frac{1-	an 	heta}{1+	an 	heta}ight] d 	heta = \int_{0}^{\pi/4} \log \left(\frac{2}{1+	an 	heta}ight) d 	heta$$I = \int_{0}^{\pi/4} [\log 2 - \log (1+	an 	heta)] d 	heta \quad ...(ii)$Adding $(i)$ and $(ii)$:$2I = \int_{0}^{\pi/4} \log 2 d 	heta = \log 2 [	heta]_{0}^{\pi/4} = \frac{\pi}{4} \log 2$.Therefore,$I = \frac{\pi}{8} \log 2$.

Find the value of $\int_{0}^{1} \frac{\log (1+x)}{1+x^{2}} d x$.

Similar Questions

$\int_0^{\pi / 2} \frac{200 \sin x+100 \cos x}{\sin x+\cos x} d x$ is equal to (in $\pi$)

Evaluate the definite integral: $\int_{0}^{\infty} \frac{x}{(1 + x)(1 + x^2)} \, dx$

By using the properties of definite integrals,evaluate the integral $\int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}} d x$.

If $n$ is any integer,then $\int_0^\pi {e^{\cos^2 x} \cos^3((2n + 1)x)} \, dx = $

The value of the integral $\int_0^{\frac{\pi}{2}} \frac{\sqrt{\cot x}}{\sqrt{\cot x}+\sqrt{\tan x}} \,dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module