Integrals - Evaluation of Definite Integrals by Substitution

Evaluate $\int_{0}^{1} \frac{\tan^{-1} x}{1 + x^2} dx$

Let $u = \tan^{-1} x$. Then $du = \frac{1}{1 + x^2} dx$.

Change of limits: When $x = 0$, $u = \tan^{-1}(0) = 0$. When $x = 1$, $u = \tan^{-1}(1) = \frac{\pi}{4}$.

The integral becomes: $\int_{0}^{\pi/4} u \, du = \left[ \frac{u^2}{2} \right]_{0}^{\pi/4}$ $= \frac{1}{2} \left( \left(\frac{\pi}{4}\right)^2 - 0^2 \right)$ $= \frac{1}{2} \cdot \frac{\pi^2}{16} = \frac{\pi^2}{32}$

We use the substitution $u = \tan^{-1} x$ because its derivative $\frac{1}{1+x^2}$ is present in the integrand. We change the limits from $x \in [0, 1]$ to $u \in [0, \frac{\pi}{4}]$ and evaluate directly.

Evaluate $\int_{2}^{3} \frac{x}{x^2 + 1} dx$

Let $u = x^2 + 1$. Then $du = 2x \, dx$, which implies $x \, dx = \frac{1}{2} du$.

Change of limits: When $x = 2$, $u = 2^2 + 1 = 5$. When $x = 3$, $u = 3^2 + 1 = 10$.

The integral becomes: $\int_{5}^{10} \frac{1}{u} \cdot \frac{1}{2} du = \frac{1}{2} [ \log |u| ]_{5}^{10}$ $= \frac{1}{2} (\log 10 - \log 5)$ $= \frac{1}{2} \log\left(\frac{10}{5}\right) = \frac{1}{2} \log 2$

By substituting the denominator $u = x^2 + 1$, the numerator becomes a part of $du$. The limits are updated to $5$ and $10$, and the standard integral $\int \frac{1}{u} du = \log |u|$ is used.