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

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

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integral of (x^3)/(a^2-x^2)

\int \frac{x ^{3}}{a ^{2} - x ^{2}} d x

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

Solution steps

\int \frac{x ^{3}}{a ^{2} - x ^{2}} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

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

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

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

\int 1 d u = u

\int \frac{a ^{2}}{u} d u = a^{2} ln ∣ u ∣

= - \frac{1}{2} (- u + a^{2} ln ∣ u ∣)

Substitute back

u = a^{2} - x^{2}

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

Simplify

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

Add a constant to the solution

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

\frac{t ^{5} - 3 t ^{2}}{2 t}

\int \frac{x}{x + 5} d x

\int (x - 2) (3 x - 4) d x

\int 3 y sin (x y) d x

\frac{d y}{d x} = 7 x^{5} + 6

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