Integration of a function involving ln: $\int \frac{x}{x^2 + 2} dx$

Question

Given the function,

$\dfrac{x}{x^2 + 2} $

The integral of the above term produces,

$\frac{1}{2}\ln|x^2 +2| + k$

I do not know how to reach this as I would have given the following term as an integral term,

$2x \ln|x^2 +2|$

Answer 1

It might help you to use $u$-substitution. We have: $$\int \frac x{x^2 + 2} \,dx$$

If we let $u= x^2 + 2 $, then $du = 2x\,dx\;$ or $\;x dx = \frac 12 du$

That gives us $$\frac 12 \int \frac 1u\,du = \frac 12 \ln|u| + c = \frac 12 \ln|x^2 +2| + c$$

Notice that, although $\int \dfrac 1x \,dx = \ln |x| + c$, when we have a function of $x$, say $f(x)$ in the denominator, then we need to have $$\int \frac{f'(x)}{f(x)}\,dx$$ in order to integrate and obtain $\ln|f(x)| + c$

Answer 2

Hint. One may observe that $$ \int \frac{x}{x^2+2}\:dx=\frac12\int \frac{2x}{x^2+2}\:dx=\frac12\int \frac{(x^2+2)'}{(x^2+2)}\:dx. $$

Answer 3

So you look for $$ \int \frac{x}{x^2 + 2} dx $$ Do the change of variable : $t = x^2 + 2, \ dt = 2x\; dx$ then : $$ \int \frac{x}{x^2 + 2} dx = \int \frac{dt / 2}{t} = \frac{1}{2} \ln(t) $$

Answer 4

Generally, for any differentiable $f\neq 0$ and some constant $a\neq 0$:

$$\int \dfrac{f'(x)}{af(x)} \mathrm{d}x = \frac{1}{a} \ln \mid f(x) \mid + C$$

where $C$ is the constant of integration.