What is the integral of 36t(9t^2-7)^3 ?

The integral of 36t(9t^2-7)^3 is 1/2 (9t^2-7)^4+C

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integral of 36t(9t^2-7)^3

\int 36 t (9 t^{2} - 7)^{3} d t

\frac{1}{2} (9 t^{2} - 7)^{4} + C

Solution steps

\int 36 t (9 t^{2} - 7)^{3} d t

Take the constant out:

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

= 36 \cdot \int t (9 t^{2} - 7)^{3} d t

Apply u-substitution

: \int \frac{u ^{3}}{18} d u

= 36 \cdot \int \frac{u ^{3}}{18} d u

Take the constant out:

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

= 36 \cdot \frac{1}{18} \cdot \int u^{3} d u

Apply the Power Rule

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

= 36 \cdot \frac{1}{18} \cdot \frac{u ^{4}}{4}

Substitute back

u = 9 t^{2} - 7

= 36 \cdot \frac{1}{18} \cdot \frac{( 9 t ^{2} - 7 ) ^{4}}{4}

Simplify

36 \cdot \frac{1}{18} \cdot \frac{( 9 t ^{2} - 7 ) ^{4}}{4} : \frac{1}{2} (9 t^{2} - 7)^{4}

= \frac{1}{2} (9 t^{2} - 7)^{4}

Add a constant to the solution

= \frac{1}{2} (9 t^{2} - 7)^{4} + C

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What is the integral of 36t(9t^2-7)^3 ?
The integral of 36t(9t^2-7)^3 is 1/2 (9t^2-7)^4+C