What is the integral from x to x^2 of 1/(t^3+2) ?

The integral from x to x^2 of 1/(t^3+2) is

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integral from x to x^2 of 1/(t^3+2)

\int_{x}^{x^{2}} \frac{1}{t ^{3} + 2} d t

Solution steps

\int_{x}^{x^{2}} \frac{1}{t ^{3} + 2} d t

t^{3} + 2 = 2 (\frac{t ^{3}}{2} + 1)

= \int_{x}^{x^{2}} \frac{1}{2 ( \frac{t ^{3}}{2} + 1 )} d t

Apply exponent rule:

\frac{a ^{n}}{b ^{n}} = (\frac{a}{b})^{n}

Take the constant out:

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

Apply u-substitution

Take the constant out:

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

Take the partial fraction of

\frac{1}{u ^{3} + 1} : \frac{1}{3 ( u + 1 )} + \frac{- u + 2}{3 ( u ^{2} - u + 1 )}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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What is the integral from x to x^2 of 1/(t^3+2) ?
The integral from x to x^2 of 1/(t^3+2) is