What is the integral of (x^3+4)/(x^2+4) ?

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

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integral of (x^3+4)/(x^2+4)

\int \frac{x ^{3} + 4}{x ^{2} + 4} d x

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

Solution steps

\int \frac{x ^{3} + 4}{x ^{2} + 4} d x

Apply Integral Substitution

= \int \frac{2 ( 2 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

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

Long division

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

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

Apply the Sum Rule:

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

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

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

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

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

Substitute back

u = \frac{x}{2}

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (cos (x - y) + (x + y))

y = x, y = 2 x 25 - x^{2}

\int_{- 2}^{2} - 2 x^{2} + 8 d x

f (x) = \frac{- 2}{x ^{2}}

\frac{1}{x ^{2} + 18 x + 88}

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