What is the integral of 1/((x^2+1)^{1.5)} ?

The integral of 1/((x^2+1)^{1.5)} is x/(sqrt(1+x^2))+C

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integral of 1/((x^2+1)^{1.5)}

\int \frac{1}{( x ^{2} + 1 ) ^{1.5}} d x

\frac{x}{1 + x ^{2}} + C

Solution steps

\int \frac{1}{( x ^{2} + 1 ) ^{1.5}} d x

Apply Trigonometric Substitution

: \int cos (u) d u

= \int cos (u) d u

Use the common integral:

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

= sin (u)

Substitute back

u = arctan (x)

= sin (arctan (x))

Use the following identity:

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

= \frac{x}{1 + x ^{2}}

Add a constant to the solution

= \frac{x}{1 + x ^{2}} + C

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What is the integral of 1/((x^2+1)^{1.5)} ?
The integral of 1/((x^2+1)^{1.5)} is x/(sqrt(1+x^2))+C