What is the integral of (3t+8)^{2.8} ?

The integral of (3t+8)^{2.8} is 0.08771…(3t+8)^{3.8}+C

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integral of (3t+8)^{2.8}

\int (3 t + 8)^{2.8} d t

0.08771 \dots (3 t + 8)^{3.8} + C

Solution steps

\int (3 t + 8)^{2.8} d t

Apply u-substitution

: \int \frac{u ^{2.8}}{3} d u

= \int \frac{u ^{2.8}}{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^{2.8} d u

Apply the Power Rule

: 0.26315 \dots u^{3.8}

= \frac{1}{3} \cdot 0.26315 \dots u^{3.8}

Substitute back

u = 3 t + 8

= \frac{1}{3} \cdot 0.26315 \dots (3 t + 8)^{3.8}

Simplify

\frac{1}{3} \cdot 0.26315 \dots (3 t + 8)^{3.8} : 0.08771 \dots (3 t + 8)^{3.8}

= 0.08771 \dots (3 t + 8)^{3.8}

Add a constant to the solution

= 0.08771 \dots (3 t + 8)^{3.8} + C

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What is the integral of (3t+8)^{2.8} ?
The integral of (3t+8)^{2.8} is 0.08771…(3t+8)^{3.8}+C