What is the integral of (1/(sqrt(x^2+1)))^3 ?

The integral of (1/(sqrt(x^2+1)))^3 is x/(sqrt(1+x^2))+C

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integral of (1/(sqrt(x^2+1)))^3

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

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

Solution steps

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

Apply radical rule:

a = a^{\frac{1}{2}}

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

= \int ((x^{2} + 1)^{- \frac{1}{2}})^{3} d x

Simplify

((x^{2} + 1)^{- \frac{1}{2}})^{3} : \frac{1}{( x ^{2} + 1 ) ^{\frac{3}{2}}}

= \int \frac{1}{( x ^{2} + 1 ) ^{\frac{3}{2}}} 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

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

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\int \frac{3}{x + 5} d x

\int (\frac{4}{x ^{5}} - \frac{5}{x ^{2}} + \frac{7}{x}) d x

What is the integral of (1/(sqrt(x^2+1)))^3 ?
The integral of (1/(sqrt(x^2+1)))^3 is x/(sqrt(1+x^2))+C