What is the integral of (t^2)/((t^3-2)^5) ?

The integral of (t^2)/((t^3-2)^5) is -1/(12(t^3-2)^4)+C

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integral of (t^2)/((t^3-2)^5)

\int \frac{t ^{2}}{( t ^{3} - 2 ) ^{5}} d t

- \frac{1}{12 ( t ^{3} - 2 ) ^{4}} + C

Solution steps

\int \frac{t ^{2}}{( t ^{3} - 2 ) ^{5}} d t

Apply u-substitution

: \int \frac{1}{3 u ^{5}} d u

= \int \frac{1}{3 u ^{5}} d u

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}{u ^{5}} d u

Apply the Power Rule

: - \frac{1}{4 u ^{4}}

= \frac{1}{3} (- \frac{1}{4 u ^{4}})

Substitute back

u = t^{3} - 2

= \frac{1}{3} (- \frac{1}{4 ( t ^{3} - 2 ) ^{4}})

Simplify

\frac{1}{3} (- \frac{1}{4 ( t ^{3} - 2 ) ^{4}}) : - \frac{1}{12 ( t ^{3} - 2 ) ^{4}}

= - \frac{1}{12 ( t ^{3} - 2 ) ^{4}}

Add a constant to the solution

= - \frac{1}{12 ( t ^{3} - 2 ) ^{4}} + C

\frac{d}{d x} (x^{0.85})

n = 1 \sum \infty (1 - \frac{1}{3 n})^{n}

x \to - 3 lim (\frac{x ^{2} - 6}{3 - x})

y^{^{''}} - 6 y^{^{'}} + 36 y = 0

\int \frac{1}{w ^{6}} d w

What is the integral of (t^2)/((t^3-2)^5) ?
The integral of (t^2)/((t^3-2)^5) is -1/(12(t^3-2)^4)+C