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

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

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

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

\frac{1}{7 ( 1 - 7 t ) ^{8}} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int - \frac{1}{7 u ^{9}} d u

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

Take the constant out:

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

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

Apply the Power Rule

: - \frac{1}{8 u ^{8}}

= 8 (- \frac{1}{7} (- \frac{1}{8 u ^{8}}))

Substitute back

u = 1 - 7 t

= 8 (- \frac{1}{7} (- \frac{1}{8 ( 1 - 7 t ) ^{8}}))

Simplify

8 (- \frac{1}{7} (- \frac{1}{8 ( 1 - 7 t ) ^{8}})) : \frac{1}{7 ( 1 - 7 t ) ^{8}}

= \frac{1}{7 ( 1 - 7 t ) ^{8}}

Add a constant to the solution

= \frac{1}{7 ( 1 - 7 t ) ^{8}} + C

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