What is the integral of ((x^3))/(sqrt(1+x^2)) ?

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

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integral of ((x^3))/(sqrt(1+x^2))

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

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

Solution steps

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

Apply u-substitution

: \int u^{2} - 1 d u

= \int u^{2} - 1 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 1 d u

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

\int 1 d u = u

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

Substitute back

u = 1 + x^{2}

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

Simplify

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

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

Add a constant to the solution

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

2 x^{2} + x - 2

\int_{0}^{\infty} \frac{5}{x ^{3}} d x

0 = x^{2} - x y + 2 y - 2, at x = - 2

\frac{\partial}{\partial x} (x^{3} \cdot y^{2} + 4 x^{2} - 3 x y)

x \to \infty lim (x - {x})

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