What is the integral of 5x(2x^2+1)^{-3} ?

The integral of 5x(2x^2+1)^{-3} is -5/(8(2x^2+1)^2)+C

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integral of 5x(2x^2+1)^{-3}

\int 5 x (2 x^{2} + 1)^{- 3} d x

- \frac{5}{8 ( 2 x ^{2} + 1 ) ^{2}} + C

Solution steps

\int 5 x (2 x^{2} + 1)^{- 3} d x

Take the constant out:

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

= 5 \cdot \int x (2 x^{2} + 1)^{- 3} d x

Simplify

x (2 x^{2} + 1)^{- 3} : \frac{x}{( 2 x ^{2} + 1 ) ^{3}}

= 5 \cdot \int \frac{x}{( 2 x ^{2} + 1 ) ^{3}} d x

Apply u-substitution

= 5 \cdot \int \frac{1}{4 u ^{3}} d u

Take the constant out:

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

= 5 \cdot \frac{1}{4} \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

= 5 \cdot \frac{1}{4} (- \frac{1}{2 u ^{2}})

Substitute back

u = 2 x^{2} + 1

= 5 \cdot \frac{1}{4} (- \frac{1}{2 ( 2 x ^{2} + 1 ) ^{2}})

Simplify

5 \cdot \frac{1}{4} (- \frac{1}{2 ( 2 x ^{2} + 1 ) ^{2}}) : - \frac{5}{8 ( 2 x ^{2} + 1 ) ^{2}}

= - \frac{5}{8 ( 2 x ^{2} + 1 ) ^{2}}

Add a constant to the solution

= - \frac{5}{8 ( 2 x ^{2} + 1 ) ^{2}} + C

What is the integral of 5x(2x^2+1)^{-3} ?
The integral of 5x(2x^2+1)^{-3} is -5/(8(2x^2+1)^2)+C