What is the integral of 3t^2(t^3+4)^5 ?

The integral of 3t^2(t^3+4)^5 is 1/6 (t^3+4)^6+C

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integral of 3t^2(t^3+4)^5

\int 3 t^{2} (t^{3} + 4)^{5} d t

\frac{1}{6} (t^{3} + 4)^{6} + C

Solution steps

\int 3 t^{2} (t^{3} + 4)^{5} d t

Take the constant out:

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

= 3 \cdot \int t^{2} (t^{3} + 4)^{5} d t

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

: \frac{u ^{6}}{6}

= 3 \cdot \frac{1}{3} \cdot \frac{u ^{6}}{6}

Substitute back

u = t^{3} + 4

= 3 \cdot \frac{1}{3} \cdot \frac{( t ^{3} + 4 ) ^{6}}{6}

Simplify

3 \cdot \frac{1}{3} \cdot \frac{( t ^{3} + 4 ) ^{6}}{6} : \frac{1}{6} (t^{3} + 4)^{6}

= \frac{1}{6} (t^{3} + 4)^{6}

Add a constant to the solution

= \frac{1}{6} (t^{3} + 4)^{6} + C

\frac{1}{1 + \frac{1}{x}}, 0

\int_{- 1}^{1} (x - 1 - x^{2}) d x

t y^{^{'}} = 3 y + t^{4} sin (4 t)

P^{^{'}} = 200 e^{0.02 t}

f (x) = x + 5

What is the integral of 3t^2(t^3+4)^5 ?
The integral of 3t^2(t^3+4)^5 is 1/6 (t^3+4)^6+C