Evaluate the definite integral int_{0}^{1} fr | vedclass

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

Question

(A) To evaluate the integral $\int_{0}^{1} \frac{x}{x^{2}+1} d x$,we use the method of substitution.Let $x^{2}+1 = t$. Then,differentiating both sides

Vedclass · Accepted Answer

$\frac{1}{2} \log 2$

Vedclass · Answer

(A) To evaluate the integral $\int_{0}^{1} \frac{x}{x^{2}+1} d x$,we use the method of substitution.Let $x^{2}+1 = t$. Then,differentiating both sides with respect to $x$,we get $2x dx = dt$,which implies $x dx = \frac{1}{2} dt$.Now,change the limits of integration:When $x = 0$,$t = 0^{2} + 1 = 1$.When $x = 1$,$t = 1^{2} + 1 = 2$.Substituting these into the integral:$\int_{0}^{1} \frac{x}{x^{2}+1} d x = \int_{1}^{2} \frac{1}{t} \cdot \frac{1}{2} dt = \frac{1}{2} \int_{1}^{2} \frac{1}{t} dt$.The integral of $\frac{1}{t}$ is $\log |t|$.$= \frac{1}{2} [\log |t|]_{1}^{2}$.$= \frac{1}{2} (\log 2 - \log 1)$.Since $\log 1 = 0$,we get:$= \frac{1}{2} \log 2$.

Evaluate the definite integral $\int_{0}^{1} \frac{x}{x^{2}+1} d x$.

Similar Questions

Evaluate the following integral:
$\int_{4}^{9} \frac{\sqrt{x}}{\left(30-x^{\frac{3}{2}}\right)^{2}} d x$

Evaluate the integral $\int_{0}^{1} \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x$.

Evaluate the integral $\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^{2} x} \,d x$.

The value of the integral $\int_0^{\frac{1}{2}} \frac{1+\sqrt{3}}{\left((x+1)^2(1-x)^6\right)^{\frac{1}{4}}} d x$ is . . . . . . . .

$\int_{\log _e 2}^x \frac{d t}{\sqrt{e^t-1}}=\frac{\pi}{6} \Rightarrow x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module