What is the integral of 9/((7x+4)^5) ?

The integral of 9/((7x+4)^5) is -9/(28(7x+4)^4)+C

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integral of 9/((7x+4)^5)

\int \frac{9}{( 7 x + 4 ) ^{5}} d x

- \frac{9}{28 ( 7 x + 4 ) ^{4}} + C

Solution steps

\int \frac{9}{( 7 x + 4 ) ^{5}} d x

Take the constant out:

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

= 9 \cdot \int \frac{1}{( 7 x + 4 ) ^{5}} d x

Apply u-substitution

: \int \frac{1}{7 u ^{5}} d u

= 9 \cdot \int \frac{1}{7 u ^{5}} d u

Take the constant out:

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

= 9 \cdot \frac{1}{7} \cdot \int \frac{1}{u ^{5}} d u

Apply the Power Rule

: - \frac{1}{4 u ^{4}}

= 9 \cdot \frac{1}{7} (- \frac{1}{4 u ^{4}})

Substitute back

u = 7 x + 4

= 9 \cdot \frac{1}{7} (- \frac{1}{4 ( 7 x + 4 ) ^{4}})

Simplify

9 \cdot \frac{1}{7} (- \frac{1}{4 ( 7 x + 4 ) ^{4}}) : - \frac{9}{28 ( 7 x + 4 ) ^{4}}

= - \frac{9}{28 ( 7 x + 4 ) ^{4}}

Add a constant to the solution

= - \frac{9}{28 ( 7 x + 4 ) ^{4}} + C

\int 2 x^{2} cos (5 x) d x

y^{^{'}} + 7 y = 5 cos (2 t)

y^{^{''}} - 2 y^{^{'}} + 2 y = 2 x - 2

x \to 0 + lim (2^{- \frac{1}{x}})

\int \frac{1}{( x ^{2} + y ^{2} + z ^{2} ) ^{\frac{3}{2}}} d x

What is the integral of 9/((7x+4)^5) ?
The integral of 9/((7x+4)^5) is -9/(28(7x+4)^4)+C