Question #31e26

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Answer 1 · 2016-12-04T02:14:50

$intarctan(sqrt(x))dx=(x+1)arctan(sqrt(x))-sqrt(x)+C$

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

Let $t = sqrt(x) => dt = 1/(2sqrt(x))dx$ . Then

$intarctan(sqrt(x))dx = int(2sqrt(x)arctan(sqrt(x)))/(2sqrt(x))dx$

$=int2tarctan(t)dt$

Next, we will apply integration by parts. To apply the formula $intudv = uv-intvdu$ , we let

$u = arctan(t)$ and $dv = 2tdt$
$=> du = 1/(1+t^2)dt$ and $v = t^2$

Applying the formula, this gives

$int2tarctan(t)dt = t^2arctan(t)-intt^2/(1+t^2)dt$

Focusing on the remaining integral, we have

$intt^2/(1+t^2)dt = int(1-1/(1+t^2))dt$

$=intdt - int1/(1+t^2)dt$

$=t - arctan(t) + C$

Putting this all together, we get our final result:

$intarctan(sqrt(x))dx = int2tarctan(t)dt$

$= t^2arctan(t)-intt^2/(1+t^2)dt$

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

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

$=(x+1)arctan(sqrt(x))-sqrt(x)+C$