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

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

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integral of (3t^4-8)t^3

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

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

Solution steps

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

Apply Integration By Parts

: \frac{1}{4} t^{4} (3 t^{4} - 8) - \int 3 t^{7} d t

= \frac{1}{4} t^{4} (3 t^{4} - 8) - \int 3 t^{7} d t

\int 3 t^{7} d t = \frac{3 t ^{8}}{8}

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

Add a constant to the solution

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

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