What is the integral of (10x)/((6-5x^2)^4) ?

The integral of (10x)/((6-5x^2)^4) is -1/(3(-6+5x^2)^3)+C

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integral of (10x)/((6-5x^2)^4)

\int \frac{10 x}{( 6 - 5 x ^{2} ) ^{4}} d x

- \frac{1}{3 ( - 6 + 5 x ^{2} ) ^{3}} + C

Solution steps

\int \frac{10 x}{( 6 - 5 x ^{2} ) ^{4}} d x

Take the constant out:

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

= 10 \cdot \int \frac{x}{( 6 - 5 x ^{2} ) ^{4}} d x

Factor

6 - 5 x^{2} : - (- 6 + 5 x^{2})

= 10 \cdot \int \frac{x}{( - ( - 6 + 5 x ^{2} ) ) ^{4}} d x

(- (- 6 + 5 x^{2}))^{4} = (- 6 + 5 x^{2})^{4}

= 10 \cdot \int \frac{x}{( - 6 + 5 x ^{2} ) ^{4}} d x

Apply u-substitution

: \int \frac{1}{10 u ^{4}} d u

= 10 \cdot \int \frac{1}{10 u ^{4}} d u

Take the constant out:

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

= 10 \cdot \frac{1}{10} \cdot \int \frac{1}{u ^{4}} d u

Apply the Power Rule

: - \frac{1}{3 u ^{3}}

= 10 \cdot \frac{1}{10} (- \frac{1}{3 u ^{3}})

Substitute back

u = - 6 + 5 x^{2}

= 10 \cdot \frac{1}{10} (- \frac{1}{3 ( - 6 + 5 x ^{2} ) ^{3}})

Simplify

10 \cdot \frac{1}{10} (- \frac{1}{3 ( - 6 + 5 x ^{2} ) ^{3}}) : - \frac{1}{3 ( - 6 + 5 x ^{2} ) ^{3}}

= - \frac{1}{3 ( - 6 + 5 x ^{2} ) ^{3}}

Add a constant to the solution

= - \frac{1}{3 ( - 6 + 5 x ^{2} ) ^{3}} + C

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What is the integral of (10x)/((6-5x^2)^4) ?
The integral of (10x)/((6-5x^2)^4) is -1/(3(-6+5x^2)^3)+C