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

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

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

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

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

Solution steps

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

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

assuming

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

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

x \to - 3 lim (x^{2} + 7 x + 12)

\int \frac{x}{5 x ^{2} + 3} d x

\frac{d}{d x} (\frac{6 - sec ( x )}{tan ( x )})

\frac{\partial}{\partial x} (- 6 y^{3} sin (6 x))

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