What is the integral of (17x^3)/(sqrt(3+x^2)) ?

The integral of (17x^3)/(sqrt(3+x^2)) is 17(1/3 (3+x^2)^{3/2}-3sqrt(3+x^2))+C

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integral of (17x^3)/(sqrt(3+x^2))

\int \frac{17 x ^{3}}{3 + x ^{2}} d x

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

Solution steps

\int \frac{17 x ^{3}}{3 + x ^{2}} d x

Take the constant out:

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

= 17 \cdot \int \frac{x ^{3}}{3 + x ^{2}} d x

Apply u-substitution

: \int u^{2} - 3 d u

= 17 \cdot \int u^{2} - 3 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

= 17 (\int u^{2} d u - \int 3 d u)

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

\int 3 d u = 3 u

= 17 (\frac{u ^{3}}{3} - 3 u)

Substitute back

u = 3 + x^{2}

= 17 (\frac{( 3 + x ^{2} ) ^{3}}{3} - 3 3 + x^{2})

Simplify

17 (\frac{( 3 + x ^{2} ) ^{3}}{3} - 3 3 + x^{2}) : 17 (\frac{1}{3} (3 + x^{2})^{\frac{3}{2}} - 3 3 + x^{2})

= 17 (\frac{1}{3} (3 + x^{2})^{\frac{3}{2}} - 3 3 + x^{2})

Add a constant to the solution

= 17 (\frac{1}{3} (3 + x^{2})^{\frac{3}{2}} - 3 3 + x^{2}) + C

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What is the integral of (17x^3)/(sqrt(3+x^2)) ?
The integral of (17x^3)/(sqrt(3+x^2)) is 17(1/3 (3+x^2)^{3/2}-3sqrt(3+x^2))+C