What is the integral of (x^3)/(sqrt(9x^2+49)) ?

The integral of (x^3)/(sqrt(9x^2+49)) is 1/81 (1/3 (9x^2+49)^{3/2}-49sqrt(9x^2+49))+C

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integral of (x^3)/(sqrt(9x^2+49))

\int \frac{x ^{3}}{9 x ^{2} + 49} d x

\frac{1}{81} (\frac{1}{3} (9 x^{2} + 49)^{\frac{3}{2}} - 49 9 x^{2} + 49) + C

Solution steps

\int \frac{x ^{3}}{9 x ^{2} + 49} d x

Apply u-substitution

: \int \frac{u ^{2} - 49}{81} d u

= \int \frac{u ^{2} - 49}{81} d u

Take the constant out:

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

= \frac{1}{81} \cdot \int u^{2} - 49 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}{81} (\int u^{2} d u - \int 49 d u)

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

\int 49 d u = 49 u

= \frac{1}{81} (\frac{u ^{3}}{3} - 49 u)

Substitute back

u = 9 x^{2} + 49

= \frac{1}{81} (\frac{( 9 x ^{2} + 49 ) ^{3}}{3} - 49 9 x^{2} + 49)

Simplify

\frac{1}{81} (\frac{( 9 x ^{2} + 49 ) ^{3}}{3} - 49 9 x^{2} + 49) : \frac{1}{81} (\frac{1}{3} (9 x^{2} + 49)^{\frac{3}{2}} - 49 9 x^{2} + 49)

= \frac{1}{81} (\frac{1}{3} (9 x^{2} + 49)^{\frac{3}{2}} - 49 9 x^{2} + 49)

Add a constant to the solution

= \frac{1}{81} (\frac{1}{3} (9 x^{2} + 49)^{\frac{3}{2}} - 49 9 x^{2} + 49) + C

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What is the integral of (x^3)/(sqrt(9x^2+49)) ?
The integral of (x^3)/(sqrt(9x^2+49)) is 1/81 (1/3 (9x^2+49)^{3/2}-49sqrt(9x^2+49))+C