What is the integral of x/((6x^2+5)^3) ?

The integral of x/((6x^2+5)^3) is -1/(24(6x^2+5)^2)+C

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integral of x/((6x^2+5)^3)

\int \frac{x}{( 6 x ^{2} + 5 ) ^{3}} d x

- \frac{1}{24 ( 6 x ^{2} + 5 ) ^{2}} + C

Solution steps

\int \frac{x}{( 6 x ^{2} + 5 ) ^{3}} d x

Apply u-substitution

: \int \frac{1}{12 u ^{3}} d u

= \int \frac{1}{12 u ^{3}} d u

Take the constant out:

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

= \frac{1}{12} \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= \frac{1}{12} (- \frac{1}{2 u ^{2}})

Substitute back

u = 6 x^{2} + 5

= \frac{1}{12} (- \frac{1}{2 ( 6 x ^{2} + 5 ) ^{2}})

Simplify

\frac{1}{12} (- \frac{1}{2 ( 6 x ^{2} + 5 ) ^{2}}) : - \frac{1}{24 ( 6 x ^{2} + 5 ) ^{2}}

= - \frac{1}{24 ( 6 x ^{2} + 5 ) ^{2}}

Add a constant to the solution

= - \frac{1}{24 ( 6 x ^{2} + 5 ) ^{2}} + C

\frac{\frac{d y}{d x}}{y} = 3

\int x^{2} in x d x

n = 1 \sum \infty 8 n

f (x) = (\frac{x ^{2} + 3}{x ^{2} - 3})^{4}

n = 1 \sum \infty \frac{1}{n ^{2} + 2 n + 5}

What is the integral of x/((6x^2+5)^3) ?
The integral of x/((6x^2+5)^3) is -1/(24(6x^2+5)^2)+C