Science

Math

Humanities

... and beyond

Socratic Meta
Featured Answers

Topics

Question #07873

1 Answer

Mar 4, 2016

$int(1)/(x+sqrtx)dx = 2 ln(sqrt(x) + 1) + C$ , with $x>= 0$ and where $C$ is the constant of integration.

Explanation:

We want to compute the antiderivative of $int(1)/(x+sqrtx)dx$ .
Since we have $sqrt(x)$ , we'll assume $x>=0$ .

We will use this change of variable :
$x = phi(t) = t^2$

$int(1)/(x+sqrtx)dx$ = $int(1)/(t^2+sqrt(t^2))*dot ((t^2))dt$

$= int(2t)/(t^2+t)dt = 2 int 1/(t+1) dt = 2 ln(|t+1|) + C.$

Since $x = t^2$ , $t = sqrt(x)$ for $x >= 0$ (which was supposed above).

So $int(1)/(x+sqrtx)dx = 2 ln(|sqrt(x) + 1|) + C = 2 ln(sqrt(x) + 1) + C$ , where $C$ is the constant of integration.

Related questions

See all questions in Sigma Notation

Impact of this question

3090 views around the world

You can reuse this answer
Creative Commons License