What is the integral of x/((x^2+4)^3) ?

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

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integral of x/((x^2+4)^3)

\int \frac{x}{( x ^{2} + 4 ) ^{3}} d x

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

Solution steps

\int \frac{x}{( x ^{2} + 4 ) ^{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} + 4

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

Simplify

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

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

Add a constant to the solution

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

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

\frac{\partial}{\partial y} (x^{2} + y^{2})

\int_{0}^{8} \frac{5}{x ^{\frac{2}{3}}} d x

y = 2 x^{2} + 1

\frac{d}{d x} ((cos (3 x))^{x})

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