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

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

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

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

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

Solution steps

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

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 t

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

Simplify

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

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

Add a constant to the solution

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

cos^{2} (3 t)

y^{^{''}} + 2 y^{^{'}} + 2 y = sin (2 t)

sin^{5} (4 x^{5})

\int sin (2 x) (sin^{2} (x) + π)^{\frac{3}{2}} d x

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