What is the integral of (2t+7)^{2.9} ?

The integral of (2t+7)^{2.9} is 0.12820…(2t+7)^{3.9}+C

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integral of (2t+7)^{2.9}

\int (2 t + 7)^{2.9} d t

0.12820 \dots (2 t + 7)^{3.9} + C

Solution steps

\int (2 t + 7)^{2.9} d t

Apply u-substitution

: \int \frac{u ^{2.9}}{2} d u

= \int \frac{u ^{2.9}}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int u^{2.9} d u

Apply the Power Rule

: 0.25641 \dots u^{3.9}

= \frac{1}{2} \cdot 0.25641 \dots u^{3.9}

Substitute back

u = 2 t + 7

= \frac{1}{2} \cdot 0.25641 \dots (2 t + 7)^{3.9}

Simplify

\frac{1}{2} \cdot 0.25641 \dots (2 t + 7)^{3.9} : 0.12820 \dots (2 t + 7)^{3.9}

= 0.12820 \dots (2 t + 7)^{3.9}

Add a constant to the solution

= 0.12820 \dots (2 t + 7)^{3.9} + C

3 x^{3} - 5 x + 7 x^{4} - 9 - x^{2}

f (x) = - 3 x^{2} + 4 x + 20

x \to 3 lim (\frac{x - 5}{x + 3})

2 sin (t)

\int arctan (2 x) d x

What is the integral of (2t+7)^{2.9} ?
The integral of (2t+7)^{2.9} is 0.12820…(2t+7)^{3.9}+C