What is the integral of (x^3)/(sqrt(4x^2-36)) ?

The integral of (x^3)/(sqrt(4x^2-36)) is 1/2 (1/3 (x^2-9)^{3/2}+9sqrt(x^2-9))+C

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integral of (x^3)/(sqrt(4x^2-36))

\int \frac{x ^{3}}{4 x ^{2} - 36} d x

\frac{1}{2} (\frac{1}{3} (x^{2} - 9)^{\frac{3}{2}} + 9 x^{2} - 9) + C

Solution steps

\int \frac{x ^{3}}{4 x ^{2} - 36} d x

Apply u-substitution

: \int \frac{u ^{\frac{1}{3}}}{6 u ^{\frac{2}{3}} - 9} d u

= \int \frac{u ^{\frac{1}{3}}}{6 u ^{\frac{2}{3}} - 9} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{u ^{\frac{1}{3}}}{u ^{\frac{2}{3}} - 9} d u

Apply u-substitution

: \int \frac{3 v ^{3}}{v ^{2} - 9} d v

= \frac{1}{6} \cdot \int \frac{3 v ^{3}}{v ^{2} - 9} d v

Take the constant out:

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

= \frac{1}{6} \cdot 3 \cdot \int \frac{v ^{3}}{v ^{2} - 9} d v

Apply u-substitution

: \int w^{2} + 9 d w

= \frac{1}{6} \cdot 3 \cdot \int w^{2} + 9 d w

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}{6} \cdot 3 (\int w^{2} d w + \int 9 d w)

\int w^{2} d w = \frac{w ^{3}}{3}

\int 9 d w = 9 w

= \frac{1}{6} \cdot 3 (\frac{w ^{3}}{3} + 9 w)

Substitute back

= \frac{1}{6} \cdot 3 \frac{( ( ( x ^{3} ) ^{\frac{1}{3}} ) ^{2} - 9 ) ^{3}}{3} + 9 ((x^{3})^{\frac{1}{3}})^{2} - 9

Simplify

\frac{1}{6} \cdot 3 \frac{( ( ( x ^{3} ) ^{\frac{1}{3}} ) ^{2} - 9 ) ^{3}}{3} + 9 ((x^{3})^{\frac{1}{3}})^{2} - 9 : \frac{1}{2} (\frac{1}{3} (x^{2} - 9)^{\frac{3}{2}} + 9 x^{2} - 9)

= \frac{1}{2} (\frac{1}{3} (x^{2} - 9)^{\frac{3}{2}} + 9 x^{2} - 9)

Add a constant to the solution

= \frac{1}{2} (\frac{1}{3} (x^{2} - 9)^{\frac{3}{2}} + 9 x^{2} - 9) + C

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What is the integral of (x^3)/(sqrt(4x^2-36)) ?
The integral of (x^3)/(sqrt(4x^2-36)) is 1/2 (1/3 (x^2-9)^{3/2}+9sqrt(x^2-9))+C