What is the integral of (x^3)/(sqrt(9x^2-36)) ?

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

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integral of (x^3)/(sqrt(9x^2-36))

\int \frac{x ^{3}}{9 x ^{2} - 36} d x

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

Solution steps

\int \frac{x ^{3}}{9 x ^{2} - 36} d x

Apply u-substitution

: \int \frac{u ^{\frac{1}{3}}}{9 u ^{\frac{2}{3}} - 4} d u

= \int \frac{u ^{\frac{1}{3}}}{9 u ^{\frac{2}{3}} - 4} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int \frac{u ^{\frac{1}{3}}}{u ^{\frac{2}{3}} - 4} d u

Apply u-substitution

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

= \frac{1}{9} \cdot \int \frac{3 v ^{3}}{v ^{2} - 4} d v

Take the constant out:

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

= \frac{1}{9} \cdot 3 \cdot \int \frac{v ^{3}}{v ^{2} - 4} d v

Apply u-substitution

: \int w^{2} + 4 d w

= \frac{1}{9} \cdot 3 \cdot \int w^{2} + 4 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}{9} \cdot 3 (\int w^{2} d w + \int 4 d w)

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

\int 4 d w = 4 w

= \frac{1}{9} \cdot 3 (\frac{w ^{3}}{3} + 4 w)

Substitute back

= \frac{1}{9} \cdot 3 \frac{( ( ( x ^{3} ) ^{\frac{1}{3}} ) ^{2} - 4 ) ^{3}}{3} + 4 ((x^{3})^{\frac{1}{3}})^{2} - 4

Simplify

\frac{1}{9} \cdot 3 \frac{( ( ( x ^{3} ) ^{\frac{1}{3}} ) ^{2} - 4 ) ^{3}}{3} + 4 ((x^{3})^{\frac{1}{3}})^{2} - 4 : \frac{1}{3} (\frac{1}{3} (x^{2} - 4)^{\frac{3}{2}} + 4 x^{2} - 4)

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

Add a constant to the solution

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

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