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

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

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

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

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

Solution steps

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

Simplify

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

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

Take the partial fraction of

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

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

Add a constant to the solution

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

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

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

\int (\frac{u ^{3} + 2 u ^{2} - u}{3 u}) d u

\int \frac{3 x ^{2}}{( 196 + x ^{2} ) ^{2}} d x

\int a x^{4} + b x^{3} + 3 c d x

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