# Integration by nonobvious substitutions

The standard technique for evaluating the integral $$\int \sec x \,dx$$ is making the nonobvious substitution $$u = \sec x + \tan x, \qquad du = (\sec x \tan x + \sec^2 x) dx,$$ which transforms the integral to $$\int \frac{du}{u} .$$

What are other integrals that are nicely handled with nonobvious substitutions?

If you're unaware of a substitution, which usually happens when you have'nt attempted a problem of that particular type, the substitution usually seems non-obvious(happened to me a lot when beginning to learn integration), some of the examples which I liked the most are:

From here:$$\int \frac{\mathrm{dx}}{x^4[x(x^5-1)]^{1/3}}$$

This one's by me: Let $x^5z^3=x^5-1$. So $$x^5(z^3-1)=1\implies 5x^4(z^3-1)\mathrm{d}x+x^5(-3z^2\mathrm{d}z)=0\implies \mathrm{d}x=\frac{3xz^2\mathrm{d}z}{5(z^3-1)}$$ So: $$\int \frac{\mathrm{d}x}{x^4[x(x^5-1)]^{1/3}}\text{ or }\int x^{-13/3}(x^5-1)^{-1/3}\mathrm{d}x\\ =\int x^{-13/3}(x^5z^3)^{-1/3}.\frac{3xz^2\mathrm{d}z}{5(z^3-1)}=\frac35\int \frac{x^{-13/3}x^{-5/3}z^{-1}xz^2\mathrm{d}z}{x^{-5}}\\ =\frac35\int z\;\mathrm{d}z=\frac{3}{10}\left(\frac{x^5-1}{x^5}\right)^{2/3}+\mathcal{C}$$

Another One:

From here $$K=\int\frac{\ln x\,dx}{x^2+2x+4}$$

Try the substitution $x = \dfrac{4}{y}$ to get:

This one's by JimmyK4542 $$I = \int_{0}^{\infty}\dfrac{\ln x}{x^2+2x+4}\,dx = \int_{-\infty}^{0}\dfrac{\ln \frac{4}{y}}{\frac{16}{y^2}+\frac{8}{y}+4} \cdot \dfrac{-4}{y^2}\,dy$$

$$= \int_{0}^{\infty}\dfrac{\ln 4 - \ln y}{y^2+2y+4}\,dy = \ln 4 \int_{0}^{\infty}\dfrac{\,dy}{y^2+2y+4} - I$$

Thus, $$2I = \ln 4 \int_{0}^{\infty}\dfrac{\,dx}{x^2+2x+4}$$, which is easy to compute.