What is the integral of 8/((0.3+2V)^3) ?

The integral of 8/((0.3+2V)^3) is -2/((0.3+2V)^2)+C

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integral of 8/((0.3+2V)^3)

\int \frac{8}{( 0.3 + 2 V ) ^{3}} d V

- \frac{2}{( 0.3 + 2 V ) ^{2}} + C

Solution steps

\int \frac{8}{( 0.3 + 2 V ) ^{3}} d V

Take the constant out:

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

= 8 \cdot \int \frac{1}{( 0.3 + 2 V ) ^{3}} d V

Apply u-substitution

= 8 \cdot \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

= 8 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

= 8 \cdot \frac{1}{2} (- \frac{1}{2 u ^{2}})

Substitute back

u = 0.3 + 2 V

= 8 \cdot \frac{1}{2} (- \frac{1}{2 ( 0.3 + 2 V ) ^{2}})

Simplify

8 \cdot \frac{1}{2} (- \frac{1}{2 ( 0.3 + 2 V ) ^{2}}) : - \frac{2}{( 0.3 + 2 V ) ^{2}}

= - \frac{2}{( 0.3 + 2 V ) ^{2}}

Add a constant to the solution

= - \frac{2}{( 0.3 + 2 V ) ^{2}} + C

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What is the integral of 8/((0.3+2V)^3) ?
The integral of 8/((0.3+2V)^3) is -2/((0.3+2V)^2)+C