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

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

assuming

(x^{2} + 1) \geq 0

= \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 sec^{2} (x) (1 + tan (x)) d x

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

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

\int \frac{1 - x}{9 - 4 x ^{2}} d x

\int \frac{x}{2 - 3 x ^{2}} 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