What is the integral of-1/(3(s-1)^3) ?

The integral of-1/(3(s-1)^3) is 1/(6(s-1)^2)+C

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integral of-1/(3(s-1)^3)

\int - \frac{1}{3 ( s - 1 ) ^{3}} d s

\frac{1}{6 ( s - 1 ) ^{2}} + C

Solution steps

\int - \frac{1}{3 ( s - 1 ) ^{3}} d s

Take the constant out:

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

= - \frac{1}{3} \cdot \int \frac{1}{( s - 1 ) ^{3}} d s

Apply u-substitution

: \int \frac{1}{u ^{3}} d u

= - \frac{1}{3} \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

: - \frac{1}{2 u ^{2}}

= - \frac{1}{3} (- \frac{1}{2 u ^{2}})

Substitute back

u = s - 1

= - \frac{1}{3} (- \frac{1}{2 ( s - 1 ) ^{2}})

Simplify

- \frac{1}{3} (- \frac{1}{2 ( s - 1 ) ^{2}}) : \frac{1}{6 ( s - 1 ) ^{2}}

= \frac{1}{6 ( s - 1 ) ^{2}}

Add a constant to the solution

= \frac{1}{6 ( s - 1 ) ^{2}} + C

\int \frac{x}{x - 13} d x

\int cos^{4} (3 x) d x

y = \frac{d y}{d x} x - \frac{d y}{d x}^{2}

y = 5 x^{2}, y^{2} = \frac{1}{5} x

\frac{d y}{d x} = x^{6} y

What is the integral of-1/(3(s-1)^3) ?
The integral of-1/(3(s-1)^3) is 1/(6(s-1)^2)+C