What is the integral of 1/((1-6x)^4) ?

The integral of 1/((1-6x)^4) is 1/(18(1-6x)^3)+C

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integral of 1/((1-6x)^4)

\int \frac{1}{( 1 - 6 x ) ^{4}} d x

\frac{1}{18 ( 1 - 6 x ) ^{3}} + C

Solution steps

\int \frac{1}{( 1 - 6 x ) ^{4}} d x

Apply u-substitution

: \int - \frac{1}{6 u ^{4}} d u

= \int - \frac{1}{6 u ^{4}} d u

Take the constant out:

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

= - \frac{1}{6} \cdot \int \frac{1}{u ^{4}} d u

Apply the Power Rule

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

= - \frac{1}{6} (- \frac{1}{3 u ^{3}})

Substitute back

u = 1 - 6 x

= - \frac{1}{6} (- \frac{1}{3 ( 1 - 6 x ) ^{3}})

Simplify

- \frac{1}{6} (- \frac{1}{3 ( 1 - 6 x ) ^{3}}) : \frac{1}{18 ( 1 - 6 x ) ^{3}}

= \frac{1}{18 ( 1 - 6 x ) ^{3}}

Add a constant to the solution

= \frac{1}{18 ( 1 - 6 x ) ^{3}} + C

\frac{( s + 1 )}{( s ^{2} + s + 1 )}

x \to + 4 lim (5)

\int_{1}^{2} \frac{2 ln ( x )}{x ^{2}} d x

\frac{d}{d t} (tan (sec (cos (t))))

\int (1 + x y) d y

What is the integral of 1/((1-6x)^4) ?
The integral of 1/((1-6x)^4) is 1/(18(1-6x)^3)+C