What is the integral of (4t+9)^{2.1} ?

The integral of (4t+9)^{2.1} is 0.08064…(4t+9)^{3.1}+C

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integral of (4t+9)^{2.1}

\int (4 t + 9)^{2.1} d t

0.08064 \dots (4 t + 9)^{3.1} + C

Solution steps

\int (4 t + 9)^{2.1} d t

Apply u-substitution

: \int \frac{u ^{2.1}}{4} d u

= \int \frac{u ^{2.1}}{4} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int u^{2.1} d u

Apply the Power Rule

: 0.32258 \dots u^{3.1}

= \frac{1}{4} \cdot 0.32258 \dots u^{3.1}

Substitute back

u = 4 t + 9

= \frac{1}{4} \cdot 0.32258 \dots (4 t + 9)^{3.1}

Simplify

\frac{1}{4} \cdot 0.32258 \dots (4 t + 9)^{3.1} : 0.08064 \dots (4 t + 9)^{3.1}

= 0.08064 \dots (4 t + 9)^{3.1}

Add a constant to the solution

= 0.08064 \dots (4 t + 9)^{3.1} + C

f (x) = \frac{1}{x ^{5}}

\frac{d}{d x} (A (e^{9 x} + 9 e^{9 x} x))

n = 0 \sum \infty cos (3 nπ)

x \to 2 lim (2 x - 2)

t e^{2 t} cos (5 t)

What is the integral of (4t+9)^{2.1} ?
The integral of (4t+9)^{2.1} is 0.08064…(4t+9)^{3.1}+C