Question #31e26

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Question #31e26

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Answer 1 · 2016-12-04T02:14:50

$\int arctan (\sqrt{x}) d x = (x + 1) arctan (\sqrt{x}) - \sqrt{x} + C$ $intarctan(sqrt(x))dx=(x+1)arctan(sqrt(x))-sqrt(x)+C$

Explanation:

It is difficult to work with the $\sqrt{x}$ $sqrt(x)$ as the argument of $arctan$ $arctan$ , so we begin by making a substitution.

Let $t = \sqrt{x} \Rightarrow d t = \frac{1}{2 \sqrt{x}} d x$ $t = sqrt(x) => dt = 1/(2sqrt(x))dx$ . Then

$\int arctan (\sqrt{x}) d x = \int \frac{2 \sqrt{x} arctan (\sqrt{x})}{2 \sqrt{x}} d x$ $intarctan(sqrt(x))dx = int(2sqrt(x)arctan(sqrt(x)))/(2sqrt(x))dx$

$= \int 2 t arctan (t) d t$ $=int2tarctan(t)dt$

Next, we will apply integration by parts. To apply the formula $\int u d v = u v - \int v d u$ $intudv = uv-intvdu$ , we let

$u = arctan (t)$ $u = arctan(t)$ and $d v = 2 t d t$ $dv = 2tdt$
$\Rightarrow d u = \frac{1}{1 + t^{2}} d t$ $=> du = 1/(1+t^2)dt$ and $v = t^{2}$ $v = t^2$

Applying the formula, this gives

$\int 2 t arctan (t) d t = t^{2} arctan (t) - \int \frac{t^{2}}{1 + t^{2}} d t$ $int2tarctan(t)dt = t^2arctan(t)-intt^2/(1+t^2)dt$

Focusing on the remaining integral, we have

$\int \frac{t^{2}}{1 + t^{2}} d t = \int (1 - \frac{1}{1 + t^{2}}) d t$ $intt^2/(1+t^2)dt = int(1-1/(1+t^2))dt$

$= \int d t - \int \frac{1}{1 + t^{2}} d t$ $=intdt - int1/(1+t^2)dt$

$= t - arctan (t) + C$ $=t - arctan(t) + C$

Putting this all together, we get our final result:

$\int arctan (\sqrt{x}) d x = \int 2 t arctan (t) d t$ $intarctan(sqrt(x))dx = int2tarctan(t)dt$

$= t^{2} arctan (t) - \int \frac{t^{2}}{1 + t^{2}} d t$ $= t^2arctan(t)-intt^2/(1+t^2)dt$

$= t^{2} arctan (t) - t + arctan (t) + C$ $=t^2arctan(t) - t+arctan(t) + C$

$= (t^{2} + 1) arctan (t) - t + C$ $=(t^2+1)arctan(t) - t + C$

$= (x + 1) arctan (\sqrt{x}) - \sqrt{x} + C$ $=(x+1)arctan(sqrt(x))-sqrt(x)+C$

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Question #31e26

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