What is the integral of (x^3)/(x^4sqrt(16-x^2)) ?

The integral of (x^3)/(x^4sqrt(16-x^2)) is -1/8 (ln|(sqrt(-x^2+16))/4+1|-ln|(sqrt(-x^2+16))/4-1|)+C

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integral of (x^3)/(x^4sqrt(16-x^2))

\int \frac{x ^{3}}{x ^{4} 16 - x ^{2}} d x

- \frac{1}{8} (ln \frac{- x ^{2} + 16}{4} + 1 - ln \frac{- x ^{2} + 16}{4} - 1) + C

Solution steps

\int \frac{x ^{3}}{x ^{4} 16 - x ^{2}} d x

Simplify

\frac{x ^{3}}{x ^{4} 16 - x ^{2}} : \frac{1}{x - x ^{2} + 16}

= \int \frac{1}{x - x ^{2} + 16} d x

Apply u-substitution

: \int - \frac{1}{- u ^{2} + 16} d u

= \int - \frac{1}{- u ^{2} + 16} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{1}{- u ^{2} + 16} d u

Apply Integral Substitution

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= - \frac{1}{4} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

= - \frac{1}{4} \frac{ln \frac{- x ^{2} + 16}{4} + 1}{2} - \frac{ln \frac{- x ^{2} + 16}{4} - 1}{2}

Simplify

- \frac{1}{4} \frac{ln \frac{- x ^{2} + 16}{4} + 1}{2} - \frac{ln \frac{- x ^{2} + 16}{4} - 1}{2} : - \frac{1}{8} (ln \frac{- x ^{2} + 16}{4} + 1 - ln \frac{- x ^{2} + 16}{4} - 1)

= - \frac{1}{8} (ln \frac{- x ^{2} + 16}{4} + 1 - ln \frac{- x ^{2} + 16}{4} - 1)

Add a constant to the solution

= - \frac{1}{8} (ln \frac{- x ^{2} + 16}{4} + 1 - ln \frac{- x ^{2} + 16}{4} - 1) + C

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What is the integral of (x^3)/(x^4sqrt(16-x^2)) ?
The integral of (x^3)/(x^4sqrt(16-x^2)) is -1/8 (ln|(sqrt(-x^2+16))/4+1|-ln|(sqrt(-x^2+16))/4-1|)+C