What is the antiderivative of $\frac{x}{x^{2} + 4}$ $x/(x^2 + 4)$ ?

Question

32186 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-10T21:10:57

$\int \frac{x d x}{x^{2} + 4} = \frac{1}{2} ln (x^{2} + 4) + C$ $int (xdx)/(x^2+4) = 1/2ln(x^2+4) + C$

Substitute $t = x^{2} + 4$ $t=x^2+4$ , $d t = 2 x d x$ $dt=2xdx$ , so that:

$\int \frac{x d x}{x^{2} + 4} = \frac{1}{2} \int \frac{d t}{t} = \frac{1}{2} ln | t | + C$ $int (xdx)/(x^2+4) = 1/2 int dt/t = 1/2lnabst + C$

and undoing the substitution:

$\int \frac{x d x}{x^{2} + 4} = \frac{1}{2} ln (x^{2} + 4) + C$ $int (xdx)/(x^2+4) = 1/2ln(x^2+4) + C$

What is the antiderivative of x/(x^2 + 4)xx2+4x/(x^2 + 4)?