What is the integral of (x^3+36)/(x^2+36) ?

The integral of (x^3+36)/(x^2+36) is 6(3((x^2)/(36)+1-ln|(x^2)/(36)+1|)+arctan(x/6))+C

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integral of (x^3+36)/(x^2+36)

\int \frac{x ^{3} + 36}{x ^{2} + 36} d x

6 (3 (\frac{x ^{2}}{36} + 1 - ln \frac{x ^{2}}{36} + 1) + arctan (\frac{x}{6})) + C

Solution steps

\int \frac{x ^{3} + 36}{x ^{2} + 36} d x

Apply Integral Substitution

= \int \frac{6 ( 6 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

= 6 \cdot \int \frac{6 u ^{3} + 1}{u ^{2} + 1} d u

Expand

\frac{6 u ^{3} + 1}{u ^{2} + 1} : \frac{6 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

= 6 (\int \frac{6 u ^{3}}{u ^{2} + 1} d u + \int \frac{1}{u ^{2} + 1} d u)

\int \frac{6 u ^{3}}{u ^{2} + 1} d u = 3 (u^{2} + 1 - ln u^{2} + 1)

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

= 6 (3 (u^{2} + 1 - ln u^{2} + 1) + arctan (u))

Substitute back

u = \frac{x}{6}

= 6 (3 ((\frac{x}{6})^{2} + 1 - ln (\frac{x}{6})^{2} + 1) + arctan (\frac{x}{6}))

3 ((\frac{x}{6})^{2} + 1 - ln (\frac{x}{6})^{2} + 1) = 3 (\frac{x ^{2}}{36} + 1 - ln \frac{x ^{2}}{36} + 1)

= 6 (3 (\frac{x ^{2}}{36} + 1 - ln \frac{x ^{2}}{36} + 1) + arctan (\frac{x}{6}))

Add a constant to the solution

= 6 (3 (\frac{x ^{2}}{36} + 1 - ln \frac{x ^{2}}{36} + 1) + arctan (\frac{x}{6})) + C

\frac{d}{d x} (10 + \frac{1}{4} sin (10 π x))

\int \frac{ln ^{3} ( x )}{2} d x

y = x^{3} + 1, (2, 3)

x \to 0 lim (x cot (\frac{3}{5} x))

\frac{d}{d x} (sin (arccos (x)))

What is the integral of (x^3+36)/(x^2+36) ?
The integral of (x^3+36)/(x^2+36) is 6(3((x^2)/(36)+1-ln|(x^2)/(36)+1|)+arctan(x/6))+C