What is the integral of (t^2)/(sqrt(t^3+7)) ?

The integral of (t^2)/(sqrt(t^3+7)) is 2/3 sqrt(t^3+7)+C

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integral of (t^2)/(sqrt(t^3+7))

\int \frac{t ^{2}}{t ^{3} + 7} d t

\frac{2}{3} t^{3} + 7 + C

Solution steps

\int \frac{t ^{2}}{t ^{3} + 7} d t

Apply u-substitution

: \int \frac{2}{3} d u

= \int \frac{2}{3} d u

Integral of a constant:

\int a d x = a x

= \frac{2}{3} u

Substitute back

u = t^{3} + 7

= \frac{2}{3} t^{3} + 7

Add a constant to the solution

= \frac{2}{3} t^{3} + 7 + C

\frac{d y}{d x} + 8 y - 8 e^{- 5 x} = 0, y (0) = \frac{5}{3}

n = 1 \sum \infty \frac{4}{n ^{4}}

y = \frac{6}{x - 1}, (3, 3)

\int \frac{1}{y ^{2} + 9} d y

(14 x - x^{2})^{3}

What is the integral of (t^2)/(sqrt(t^3+7)) ?
The integral of (t^2)/(sqrt(t^3+7)) is 2/3 sqrt(t^3+7)+C