What is the integral of 1/((1+u^2)^{3/2)} ?

The integral of 1/((1+u^2)^{3/2)} is u/((1+u^2)^{1/2)}+C

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integral of 1/((1+u^2)^{3/2)}

\int \frac{1}{( 1 + u ^{2} ) ^{\frac{3}{2}}} d u

\frac{u}{( 1 + u ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( 1 + u ^{2} ) ^{\frac{3}{2}}} d u

Apply Trigonometric Substitution

= \int cos (v) d v

Use the common integral:

\int cos (v) d v = sin (v)

= sin (v)

Substitute back

v = arctan (u)

= sin (arctan (u))

Use the following identity:

sin (arctan (x)) = \frac{x}{( 1 + x ^{2} ) ^{\frac{1}{2}}}

= \frac{u}{( 1 + u ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{u}{( 1 + u ^{2} ) ^{\frac{1}{2}}} + C

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What is the integral of 1/((1+u^2)^{3/2)} ?
The integral of 1/((1+u^2)^{3/2)} is u/((1+u^2)^{1/2)}+C