How to integrate 1/(x^(3/2)+4) ?

1 Answer
Eric S.
Jun 8, 2018

Use the substitution x12=223u.

Explanation:

Let

I=∫1x32+4dx

Apply the substitution x12=223u:

I=∫14u3+4(273udu)

Simplify:

I=213∫uu3+1du

Factorize the denominator:

I=213∫u(u2−u+1)(u+1)du

Apply partial fraction decomposition:

I=2133∫(u+1u2−u+1−1u+1)du

Rearrange:

I=2136∫2u+2u2−u+1du−2133∫1u+1du

For the first term, pull out the a numerator that is the derivative of the denominator:

I=2136∫(2u−1u2−u+1−3u2−u+1)du−2133ln|u+1|

Hence

I=2136∫2u−1u2−u+1du−2132∫1u2−u+1du−2133ln|u+1|

Complete the square in the remaining term:

I=2136ln∣∣u2−u+1∣∣−213∫2(2u−1)2+3du−2133ln|u+1|

Apply the substitution 2u−1=√3tanθ:

I=2136ln∣∣u2−u+1∣∣−213√3∫dθ−2133ln|u+1|

Hence

I=2136ln∣∣u2−u+1∣∣−213√3θ−2133ln|u+1|+C

Reverse the last substitution:

I=2136ln∣∣u2−u+1∣∣−213√3tan−1(2u−1√3)−2133ln|u+1|+C

Rewrite in terms of 223u:

I=2136ln∣∣∣(223u)2−223(223u)+243∣∣∣−213√3tan−1⎛⎜⎝213(223u)−1√3⎞⎟⎠−2133ln∣∣223u+223∣∣+C

Reverse the first substitution:

I=2136ln∣∣x−223√x+243∣∣−213√3tan−1(213√x−1√3)−2133ln∣∣√x+223∣∣+C

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