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

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

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

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

= \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

\frac{\partial}{\partial x} (cos (x) cosh (y))

d er i v a t i v e x^{4} e^{x} - 5

\int 3 x^{2} \times 4 x^{4} d x

\int 5 x^{3} e^{- x^{2}} d x

d er i v a t i v e \frac{4 x + 3}{4 x - 3}

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