What is the integral of 9/((1-8t)^7) ?

The integral of 9/((1-8t)^7) is 3/(16(-8t+1)^6)+C

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integral of 9/((1-8t)^7)

\int \frac{9}{( 1 - 8 t ) ^{7}} d t

\frac{3}{16 ( - 8 t + 1 ) ^{6}} + C

Solution steps

\int \frac{9}{( 1 - 8 t ) ^{7}} d t

Take the constant out:

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

= 9 \cdot \int \frac{1}{( 1 - 8 t ) ^{7}} d t

Apply u-substitution

: \int - \frac{1}{8 u ^{7}} d u

= 9 \cdot \int - \frac{1}{8 u ^{7}} d u

Take the constant out:

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

= 9 (- \frac{1}{8} \cdot \int \frac{1}{u ^{7}} d u)

Apply the Power Rule

: - \frac{1}{6 u ^{6}}

= 9 (- \frac{1}{8} (- \frac{1}{6 u ^{6}}))

Substitute back

u = 1 - 8 t

= 9 (- \frac{1}{8} (- \frac{1}{6 ( 1 - 8 t ) ^{6}}))

Simplify

9 (- \frac{1}{8} (- \frac{1}{6 ( 1 - 8 t ) ^{6}})) : \frac{3}{16 ( - 8 t + 1 ) ^{6}}

= \frac{3}{16 ( - 8 t + 1 ) ^{6}}

Add a constant to the solution

= \frac{3}{16 ( - 8 t + 1 ) ^{6}} + C

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What is the integral of 9/((1-8t)^7) ?
The integral of 9/((1-8t)^7) is 3/(16(-8t+1)^6)+C