Question #07873

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Question #07873

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Answer 1 · 2016-03-04T21:18:24

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

Explanation:

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

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

$\int \frac{1}{x + \sqrt{x}} d x$ $int(1)/(x+sqrtx)dx$ = $\int \frac{1}{t^{2} + \sqrt{t^{2}}} \cdot . (t^{2}) d t$ $int(1)/(t^2+sqrt(t^2))*dot ((t^2))dt$

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

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

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

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Question #07873

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