What is the integral of (x^3+1)/(x^2-4) ?

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

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integral of (x^3+1)/(x^2-4)

\int \frac{x ^{3} + 1}{x ^{2} - 4} d x

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

Solution steps

\int \frac{x ^{3} + 1}{x ^{2} - 4} d x

Expand

\frac{x ^{3} + 1}{x ^{2} - 4} : \frac{x ^{3}}{x ^{2} - 4} + \frac{1}{x ^{2} - 4}

Apply the Sum Rule:

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

= \int \frac{x ^{3}}{x ^{2} - 4} d x + \int \frac{1}{x ^{2} - 4} d x

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

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

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

Add a constant to the solution

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

\int_{0}^{4} e^{- x} d x

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

f (x) = x^{4} + 5 x

\frac{d}{d x} (ln (1 + x) - x)

\int 8 w e^{7 w} d w

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