What is the integral of 9/((1-3x)^4) ?

The integral of 9/((1-3x)^4) is 1/((1-3x)^3)+C

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integral of 9/((1-3x)^4)

\int \frac{9}{( 1 - 3 x ) ^{4}} d x

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

Solution steps

\int \frac{9}{( 1 - 3 x ) ^{4}} 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}{( 1 - 3 x ) ^{4}} d x

Apply u-substitution

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

= 9 \cdot \int - \frac{1}{3 u ^{4}} d u

Take the constant out:

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

= 9 (- \frac{1}{3} \cdot \int \frac{1}{u ^{4}} d u)

Apply the Power Rule

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

= 9 (- \frac{1}{3} (- \frac{1}{3 u ^{3}}))

Substitute back

u = 1 - 3 x

= 9 (- \frac{1}{3} (- \frac{1}{3 ( 1 - 3 x ) ^{3}}))

Simplify

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

= \frac{1}{( 1 - 3 x ) ^{3}}

Add a constant to the solution

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

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What is the integral of 9/((1-3x)^4) ?
The integral of 9/((1-3x)^4) is 1/((1-3x)^3)+C