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

The integral of x/((x^2+1)^3) is -1/(4(x^2+1)^2)+C

Popular >

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

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

- \frac{1}{4 ( x ^{2} + 1 ) ^{2}} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 u ^{3}} d u

= \int \frac{1}{2 u ^{3}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

: - \frac{1}{2 u ^{2}}

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

Substitute back

u = x^{2} + 1

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

Simplify

\frac{1}{2} (- \frac{1}{2 ( x ^{2} + 1 ) ^{2}}) : - \frac{1}{4 ( x ^{2} + 1 ) ^{2}}

= - \frac{1}{4 ( x ^{2} + 1 ) ^{2}}

Add a constant to the solution

= - \frac{1}{4 ( x ^{2} + 1 ) ^{2}} + C

\frac{4.5 x - 4}{( 3 x - 4 ) ^{\frac{1}{2}}}

e^{x} + e^{x} x

\frac{d}{d x} (\frac{9}{x ^{3}} + \frac{1}{x})

\int_{0}^{1} (x^{3})^{2} d x

\frac{d y}{d x}, e^{x} y = 5

What is the integral of x/((x^2+1)^3) ?
The integral of x/((x^2+1)^3) is -1/(4(x^2+1)^2)+C