What is the integral of 7/((1-6x)^6) ?

The integral of 7/((1-6x)^6) is 7/(30(1-6x)^5)+C

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integral of 7/((1-6x)^6)

\int \frac{7}{( 1 - 6 x ) ^{6}} d x

\frac{7}{30 ( 1 - 6 x ) ^{5}} + C

Solution steps

\int \frac{7}{( 1 - 6 x ) ^{6}} d x

Take the constant out:

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

= 7 \cdot \int \frac{1}{( 1 - 6 x ) ^{6}} d x

Apply u-substitution

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

= 7 \cdot \int - \frac{1}{6 u ^{6}} d u

Take the constant out:

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

= 7 (- \frac{1}{6} \cdot \int \frac{1}{u ^{6}} d u)

Apply the Power Rule

: - \frac{1}{5 u ^{5}}

= 7 (- \frac{1}{6} (- \frac{1}{5 u ^{5}}))

Substitute back

u = 1 - 6 x

= 7 (- \frac{1}{6} (- \frac{1}{5 ( 1 - 6 x ) ^{5}}))

Simplify

7 (- \frac{1}{6} (- \frac{1}{5 ( 1 - 6 x ) ^{5}})) : \frac{7}{30 ( 1 - 6 x ) ^{5}}

= \frac{7}{30 ( 1 - 6 x ) ^{5}}

Add a constant to the solution

= \frac{7}{30 ( 1 - 6 x ) ^{5}} + C

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What is the integral of 7/((1-6x)^6) ?
The integral of 7/((1-6x)^6) is 7/(30(1-6x)^5)+C