What is the integral of x/((5x^2+2)^6) ?

The integral of x/((5x^2+2)^6) is -1/(50(5x^2+2)^5)+C

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integral of x/((5x^2+2)^6)

\int \frac{x}{( 5 x ^{2} + 2 ) ^{6}} d x

- \frac{1}{50 ( 5 x ^{2} + 2 ) ^{5}} + C

Solution steps

\int \frac{x}{( 5 x ^{2} + 2 ) ^{6}} d x

Apply u-substitution

: \int \frac{1}{2 ( 5 u + 2 ) ^{6}} d u

= \int \frac{1}{2 ( 5 u + 2 ) ^{6}} 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}{( 5 u + 2 ) ^{6}} d u

Apply u-substitution

: \int \frac{1}{5 v ^{6}} d v

= \frac{1}{2} \cdot \int \frac{1}{5 v ^{6}} d v

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{5} \cdot \int \frac{1}{v ^{6}} d v

Apply the Power Rule

: - \frac{1}{5 v ^{5}}

= \frac{1}{2} \cdot \frac{1}{5} (- \frac{1}{5 v ^{5}})

Substitute back

= \frac{1}{2} \cdot \frac{1}{5} (- \frac{1}{5 ( 5 x ^{2} + 2 ) ^{5}})

Simplify

\frac{1}{2} \cdot \frac{1}{5} (- \frac{1}{5 ( 5 x ^{2} + 2 ) ^{5}}) : - \frac{1}{50 ( 5 x ^{2} + 2 ) ^{5}}

= - \frac{1}{50 ( 5 x ^{2} + 2 ) ^{5}}

Add a constant to the solution

= - \frac{1}{50 ( 5 x ^{2} + 2 ) ^{5}} + C

\int_{0}^{1} \frac{1}{1 + x} d x

y^{^{''}} + 4 y^{^{'}} + 5 y = 5 x - 1

\frac{\partial ^{2}}{\partial x \partial y} (y^{4} cos (2 x))

\frac{\partial}{\partial y} (16 - x^{2} - 4 y)

y^{^{''}} - y = sin (2 t)

What is the integral of x/((5x^2+2)^6) ?
The integral of x/((5x^2+2)^6) is -1/(50(5x^2+2)^5)+C