What is the integral of (4t^3+8)t^2 ?

The integral of (4t^3+8)t^2 is 1/3 t^3(4t^3+8)-(2t^6)/3+C

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integral of (4t^3+8)t^2

\int (4 t^{3} + 8) t^{2} d t

\frac{1}{3} t^{3} (4 t^{3} + 8) - \frac{2 t ^{6}}{3} + C

Solution steps

\int (4 t^{3} + 8) t^{2} d t

Apply Integration By Parts

: \frac{1}{3} t^{3} (4 t^{3} + 8) - \int 4 t^{5} d t

= \frac{1}{3} t^{3} (4 t^{3} + 8) - \int 4 t^{5} d t

\int 4 t^{5} d t = \frac{2 t ^{6}}{3}

= \frac{1}{3} t^{3} (4 t^{3} + 8) - \frac{2 t ^{6}}{3}

Add a constant to the solution

= \frac{1}{3} t^{3} (4 t^{3} + 8) - \frac{2 t ^{6}}{3} + C

\frac{d}{d x} (\frac{2 x + 1}{4 x - 1})

\int \frac{( 2 x )}{x ^{2} + 1} d x

\int sin (6)

\frac{\partial}{\partial u} (sin (u) cos (v))

\frac{d y}{d x} = e^{y} sin (x)

What is the integral of (4t^3+8)t^2 ?
The integral of (4t^3+8)t^2 is 1/3 t^3(4t^3+8)-(2t^6)/3+C