What is the integral of (x^3+16)/(x^2+16) ?

The integral of (x^3+16)/(x^2+16) is 4(2((x^2)/(16)+1-ln|(x^2)/(16)+1|)+arctan(x/4))+C

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integral of (x^3+16)/(x^2+16)

\int \frac{x ^{3} + 16}{x ^{2} + 16} d x

4 (2 (\frac{x ^{2}}{16} + 1 - ln \frac{x ^{2}}{16} + 1) + arctan (\frac{x}{4})) + C

Solution steps

\int \frac{x ^{3} + 16}{x ^{2} + 16} d x

Apply Integral Substitution

= \int \frac{4 ( 4 u ^{3} + 1 )}{u ^{2} + 1} d u

Take the constant out:

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

= 4 \cdot \int \frac{4 u ^{3} + 1}{u ^{2} + 1} d u

Expand

\frac{4 u ^{3} + 1}{u ^{2} + 1} : \frac{4 u ^{3}}{u ^{2} + 1} + \frac{1}{u ^{2} + 1}

Apply the Sum Rule:

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

= 4 (\int \frac{4 u ^{3}}{u ^{2} + 1} d u + \int \frac{1}{u ^{2} + 1} d u)

\int \frac{4 u ^{3}}{u ^{2} + 1} d u = 2 (u^{2} + 1 - ln u^{2} + 1)

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= 4 (2 (u^{2} + 1 - ln u^{2} + 1) + arctan (u))

Substitute back

u = \frac{x}{4}

= 4 (2 ((\frac{x}{4})^{2} + 1 - ln (\frac{x}{4})^{2} + 1) + arctan (\frac{x}{4}))

2 ((\frac{x}{4})^{2} + 1 - ln (\frac{x}{4})^{2} + 1) = 2 (\frac{x ^{2}}{16} + 1 - ln \frac{x ^{2}}{16} + 1)

= 4 (2 (\frac{x ^{2}}{16} + 1 - ln \frac{x ^{2}}{16} + 1) + arctan (\frac{x}{4}))

Add a constant to the solution

= 4 (2 (\frac{x ^{2}}{16} + 1 - ln \frac{x ^{2}}{16} + 1) + arctan (\frac{x}{4})) + C

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What is the integral of (x^3+16)/(x^2+16) ?
The integral of (x^3+16)/(x^2+16) is 4(2((x^2)/(16)+1-ln|(x^2)/(16)+1|)+arctan(x/4))+C