What is the integral of (3t+2)^{2.4} ?

The integral of (3t+2)^{2.4} is 0.09803…(3t+2)^{3.4}+C

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integral of (3t+2)^{2.4}

\int (3 t + 2)^{2.4} d t

0.09803 \dots (3 t + 2)^{3.4} + C

Solution steps

\int (3 t + 2)^{2.4} d t

Apply u-substitution

: \int \frac{u ^{2.4}}{3} d u

= \int \frac{u ^{2.4}}{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^{2.4} d u

Apply the Power Rule

: 0.29411 \dots u^{3.4}

= \frac{1}{3} \cdot 0.29411 \dots u^{3.4}

Substitute back

u = 3 t + 2

= \frac{1}{3} \cdot 0.29411 \dots (3 t + 2)^{3.4}

Simplify

\frac{1}{3} \cdot 0.29411 \dots (3 t + 2)^{3.4} : 0.09803 \dots (3 t + 2)^{3.4}

= 0.09803 \dots (3 t + 2)^{3.4}

Add a constant to the solution

= 0.09803 \dots (3 t + 2)^{3.4} + C

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What is the integral of (3t+2)^{2.4} ?
The integral of (3t+2)^{2.4} is 0.09803…(3t+2)^{3.4}+C