What is the integral of (7x^3-3)/(x^2-4) ?

The integral of (7x^3-3)/(x^2-4) is 7/2 (x^2-4+4ln|x^2-4|)+3/4 (ln| x/2+1|-ln| x/2-1|)+C

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integral of (7x^3-3)/(x^2-4)

\int \frac{7 x ^{3} - 3}{x ^{2} - 4} d x

\frac{7}{2} (x^{2} - 4 + 4 ln x^{2} - 4) + \frac{3}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1) + C

Solution steps

\int \frac{7 x ^{3} - 3}{x ^{2} - 4} d x

Expand

\frac{7 x ^{3} - 3}{x ^{2} - 4} : \frac{7 x ^{3}}{x ^{2} - 4} - \frac{3}{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{7 x ^{3}}{x ^{2} - 4} d x - \int \frac{3}{x ^{2} - 4} d x

\int \frac{7 x ^{3}}{x ^{2} - 4} d x = \frac{7}{2} (x^{2} - 4 + 4 ln x^{2} - 4)

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

= \frac{7}{2} (x^{2} - 4 + 4 ln x^{2} - 4) - (- \frac{3}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1))

Simplify

= \frac{7}{2} (x^{2} - 4 + 4 ln x^{2} - 4) + \frac{3}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1)

Add a constant to the solution

= \frac{7}{2} (x^{2} - 4 + 4 ln x^{2} - 4) + \frac{3}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1) + C

\frac{d}{d x} (x^{3} - 7 x + 6)

f (x) = ln (289 sin^{2} (x))

\frac{d}{d x} (2 x^{4} y - 4)

y^{^{''''}} - 16 y = 0

\frac{d}{d t} (2 t^{\frac{4}{3}} - 3 t^{\frac{2}{3}})

What is the integral of (7x^3-3)/(x^2-4) ?
The integral of (7x^3-3)/(x^2-4) is 7/2 (x^2-4+4ln|x^2-4|)+3/4 (ln| x/2+1|-ln| x/2-1|)+C