What is the integral of 8/((4x+5)^3) ?

The integral of 8/((4x+5)^3) is -1/((4x+5)^2)+C

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integral of 8/((4x+5)^3)

\int \frac{8}{( 4 x + 5 ) ^{3}} d x

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

Solution steps

\int \frac{8}{( 4 x + 5 ) ^{3}} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{( 4 x + 5 ) ^{3}} d x

Apply u-substitution

: \int \frac{1}{4 u ^{3}} d u

= 8 \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

= 8 \cdot \frac{1}{4} \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= 8 \cdot \frac{1}{4} (- \frac{1}{2 u ^{2}})

Substitute back

u = 4 x + 5

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

Simplify

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

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

Add a constant to the solution

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

\frac{1}{1 + sin ( x )}

\frac{1}{( 1 - 0.8 \cdot cos ^{2} ( x ) )}

\frac{5}{( s ^{2} + 5 )}

\int \frac{4 ln ^{2} ( x ) + 49}{x} d x

f (x) = - x^{2} + 4 x, at x = 4

What is the integral of 8/((4x+5)^3) ?
The integral of 8/((4x+5)^3) is -1/((4x+5)^2)+C