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

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

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

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

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} ((4 - x^{2})^{2})

e^{\frac{z}{( z - 3 )}}

9 y^{^{''}} + 18 y^{^{'}} + 11 y = 0

\int_{1}^{2} x 2^{x^{2}} d x

y^{^{''}} + 225 y = sec (15 x), y (0) = 5, y^{^{'}} (0) = - 5

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