What is the integral of (t^2+4t)(t^3+6t^2)^6 ?

The integral of (t^2+4t)(t^3+6t^2)^6 is 1/21 (t^3+6t^2)^7+C

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integral of (t^2+4t)(t^3+6t^2)^6

\int (t^{2} + 4 t) (t^{3} + 6 t^{2})^{6} d t

\frac{1}{21} (t^{3} + 6 t^{2})^{7} + C

Solution steps

\int (t^{2} + 4 t) (t^{3} + 6 t^{2})^{6} d t

Apply u-substitution

: \int \frac{u ^{6}}{3} d u

= \int \frac{u ^{6}}{3} 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 u^{6} d u

Apply the Power Rule

: \frac{u ^{7}}{7}

= \frac{1}{3} \cdot \frac{u ^{7}}{7}

Substitute back

u = t^{3} + 6 t^{2}

= \frac{1}{3} \cdot \frac{( t ^{3} + 6 t ^{2} ) ^{7}}{7}

Simplify

\frac{1}{3} \cdot \frac{( t ^{3} + 6 t ^{2} ) ^{7}}{7} : \frac{1}{21} (t^{3} + 6 t^{2})^{7}

= \frac{1}{21} (t^{3} + 6 t^{2})^{7}

Add a constant to the solution

= \frac{1}{21} (t^{3} + 6 t^{2})^{7} + C

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What is the integral of (t^2+4t)(t^3+6t^2)^6 ?
The integral of (t^2+4t)(t^3+6t^2)^6 is 1/21 (t^3+6t^2)^7+C