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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \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

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

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

\int 3 d u = 3 u

= \frac{u ^{3}}{3} - 3 u

Substitute back

u = 3 + x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (3^{4 x})

f (x) = x^{2} - 3, (- 4, 13)

\int \frac{x ^{2} - 1}{x ^{3} + x} d x

\int_{0}^{1} x e^{3 x^{2}} d x

tan (t) y^{^{'}} - y = 1, y (\frac{π}{6}) = 3

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