How do you integrate $int [xsqrt(x^2 + 1)] dx$ ?

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Answer 1 · 2015-06-08T00:47:41

This is of the form:

$sqrt(x^2 + a^2)$

which means you can use the substitution:

$x = atantheta$

So, let:
$x = tantheta$
$dx = sec^2thetad theta$
$sqrt(x^2+1) = sqrt(tan^2theta+1) = sqrt(sec^2theta) = sectheta$

$int xsqrt(x^2+1)dx = int tanthetasecthetasec^2thetad theta$

$= int (sec^2theta)(secthetatantheta)d theta$

Notice how you can do a second substitution here, with u-substitution.

Let:
$u = sectheta$
$du = secthetatanthetad theta$

$= int u^2du = u^3/3 + C$

$= sec^3theta/3 + C$

Since $sectheta = sqrt(x^2 + 1)$ :

$=> 1/3(x^2+1)^(3/2) + C$

How do you integrate int [xsqrt(x^2 + 1)] dx int [xsqrt(x^2 + 1)] dx ?