What is the integral of (x^3)/(sqrt(x^2+100)) ?

The integral of (x^3)/(sqrt(x^2+100)) is 1/3 (x^2+100)^{3/2}-100sqrt(x^2+100)+C

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integral of (x^3)/(sqrt(x^2+100))

\int \frac{x ^{3}}{x ^{2} + 100} d x

\frac{1}{3} (x^{2} + 100)^{\frac{3}{2}} - 100 x^{2} + 100 + C

Solution steps

\int \frac{x ^{3}}{x ^{2} + 100} d x

Apply u-substitution

: \int u^{2} - 100 d u

= \int u^{2} - 100 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 100 d u

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

\int 100 d u = 100 u

= \frac{u ^{3}}{3} - 100 u

Substitute back

u = x^{2} + 100

= \frac{( x ^{2} + 100 ) ^{3}}{3} - 100 x^{2} + 100

Simplify

\frac{( x ^{2} + 100 ) ^{3}}{3} - 100 x^{2} + 100 : \frac{1}{3} (x^{2} + 100)^{\frac{3}{2}} - 100 x^{2} + 100

= \frac{1}{3} (x^{2} + 100)^{\frac{3}{2}} - 100 x^{2} + 100

Add a constant to the solution

= \frac{1}{3} (x^{2} + 100)^{\frac{3}{2}} - 100 x^{2} + 100 + C

t \to - 2 lim (\frac{t ^{3} + 8}{t + 2})

\frac{d}{d x} (sin (arccos (3 x)))

x \to 3 - lim (3 - x)

y^{^{''}} + 2 y^{^{'}} = 1, y (0) = 1, y^{^{'}} (0) = 0

\int_{0}^{2} 4 d x

What is the integral of (x^3)/(sqrt(x^2+100)) ?
The integral of (x^3)/(sqrt(x^2+100)) is 1/3 (x^2+100)^{3/2}-100sqrt(x^2+100)+C