This post comes to show several indefinite integrals which can be solved by non-trivial substitutions.

$$\int \frac{dx}{(a^2+x^2)^2} = \frac{1}{4a^3} \cdot \left(2 \cdot \arctan \frac{x}{a} + \sin \left(2 \cdot \arctan \frac{x}{a}\right)\right), \ \ \ where \ a \gt 0 \tag{1}$$

$$\int \frac{x^3 dx}{(a^2+x^2)^3} = \frac{1}{4a^2} \cdot \sin^4\left(\arctan \frac{x}{a}\right), \ \ \ where \ a \gt 0 \tag{2}$$

The above two integrals (1) and (2) can be solved by using the substitution $x = a \cdot \tan t$.

This solution uses the fact that the tangent function is a well-known monotonic bijection from $(-\frac{\pi}{2}, \frac{\pi}{2})$ to $\mathbb{R}$.

Below are two more integrals which are solved by a different substitution.

$$\int \sqrt{a^2+x^2} dx = \frac{a^2}{2} \cdot \left(\ln\left(\frac{x}{a} + \sqrt{\frac{x^2}{a^2} + 1}\right) + \frac{x}{a} \cdot \sqrt{\frac{x^2}{a^2} + 1}\right), \ \ \ where \ a \gt 0 \tag{3}$$

$$\int \frac{dx} {(a^2+x^2)^\frac{3}{2}} = \frac{1}{a^2} \cdot \frac{x^2 + x \cdot \sqrt{x^2+a^2}}{x^2 + a^2 + x \cdot \sqrt{x^2+a^2}}, \ \ \ where \ a \gt 0 \tag{4}$$

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