How to solve $\int \frac{1}{x^2+4x+7} dx$?

Question

How to solve $\int \frac{1}{x^2+4x+7} dx$?

I think the first step is to write it in the following form: $$\int \frac{1}{(x+2)^2+3} dx$$

Answer 1

A second step could be the change of variable $$ x+2=\sqrt{3}u, \implies (x+2)^2+3=3u^2+3=3(u^2+1) $$ giving $$ \int \frac{1}{(x+2)^2+3} dx=\frac1{\sqrt{3}}\int \frac{1}{u^2+1} du $$ which is now a standard integral.

Answer 2

Continue on:

$$\int\frac{dx}{3+(x+2)^2}=\frac1{\sqrt3}\int\frac{\frac1{\sqrt3}dx}{1+\left(\frac{x+2}{\sqrt3}\right)^2}$$

and now you have an integral of the form

$$\int\frac{f'}{1+f^2}dx$$

Answer 3

With the substitution $u=\frac{x+2}{\sqrt{3}}$:

\begin{align*}\int \frac{1}{(x+2)^2+3} \mathrm{d}x &= \int \frac{1}{3(\frac{x+2}{\sqrt{3}})^2+3} \mathrm{d}x \\ &= \frac13\int \frac{1}{(\frac{x+2}{\sqrt{3}})^2+1} \mathrm{d}x \\ &= \frac{1}{3}\sqrt{3}\int \frac{1}{u^2+1} \mathrm{d}u \\ &= \frac{1}{3}\sqrt{3} \arctan(u)+C \\ &= \frac{1}{3}\sqrt{3} \arctan\left(\frac{x+2}{\sqrt{3}}\right)+C \end{align*}

Answer 4

$$\int \frac { 1 }{ (x+2)^{ 2 }+3 } dx=\int { \frac { dx }{ 3\left( 1+{ \left( \frac { x+2 }{ \sqrt { 3 } } \right) }^{ 2 } \right) } } =\frac { 1 }{ \sqrt { 3 } } \int { \frac { d\left( \frac { x+2 }{ \sqrt { 3 } } \right) }{ 1+{ \left( \frac { x+2 }{ \sqrt { 3 } } \right) }^{ 2 } } } =\\=\frac { 1 }{ \sqrt { 3 } } \arctan { { \left( \frac { x+2 }{ \sqrt { 3 } } \right) } } +C$$