Evaluate the definite integral $\int_{2}^{3} \frac{x}{x^{2}+1} dx$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) Let $I = \int_{2}^{3} \frac{x}{x^{2}+1} dx$.
To evaluate this,we use the substitution method. Let $u = x^{2} + 1$. Then $du = 2x dx$,which implies $x dx = \frac{1}{2} du$.
Changing the limits of integration:
When $x = 2$,$u = 2^{2} + 1 = 5$.
When $x = 3$,$u = 3^{2} + 1 = 10$.
Substituting these into the integral:
$I = \int_{5}^{10} \frac{1}{2u} du = \frac{1}{2} [\ln |u|]_{5}^{10}$.
Applying the fundamental theorem of calculus:
$I = \frac{1}{2} (\ln 10 - \ln 5) = \frac{1}{2} \ln \left(\frac{10}{5}\right) = \frac{1}{2} \ln 2$.

Explore More

Similar Questions

If $\frac{d[f(x)]}{dx} = g(x)$ for $a \le x \le b$,then $\int_a^b f(x)g(x) dx$ equals

$\int_0^{\pi /4} \sec^7 \theta \sin^3 \theta \, d\theta = $

$\int_{\frac{2}{e}}^{\frac{1}{e}} \frac{1}{x(\log x)^{\frac{1}{3}}} dx$ is equal to

Let $\int_\alpha^{\log _e 4} \frac{dx}{\sqrt{e^{x}-1}}=\frac{\pi}{6}$. Then $e^\alpha$ and $e^{-\alpha}$ are the roots of the equation :

$\int_0^{\pi /4} \frac{\sec^2 x}{(1 + \tan x)(2 + \tan x)} \,dx = $

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo