What is the integral of (3t+3)^{2.2} ?

The integral of (3t+3)^{2.2} is 0.10416…(3t+3)^{3.2}+C

integral of (3t+3)^{2.2}

\int (3 t + 3)^{2.2} d t

0.10416 \dots (3 t + 3)^{3.2} + C

Solution steps

\int (3 t + 3)^{2.2} d t

Apply u-substitution

= \int \frac{u ^{2.2}}{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^{2.2} d u

Apply the Power Rule

= \frac{1}{3} \cdot 0.3125 u^{3.2}

Substitute back

u = 3 t + 3

= \frac{1}{3} \cdot 0.3125 (3 t + 3)^{3.2}

Simplify

\frac{1}{3} \cdot 0.3125 (3 t + 3)^{3.2} : 0.10416 \dots (3 t + 3)^{3.2}

= 0.10416 \dots (3 t + 3)^{3.2}

Add a constant to the solution

= 0.10416 \dots (3 t + 3)^{3.2} + C

y = 6 x - 11

\int (\frac{5}{x} + sec^{2} (x)) d x

x \to 2 lim (3.5 + 5.6)

x \to 2 - lim (\frac{2 - ∣ x ∣}{2 + x})

(4 x + 3)^{5}

What is the integral of (3t+3)^{2.2} ?
The integral of (3t+3)^{2.2} is 0.10416…(3t+3)^{3.2}+C