What is the integral of (9x^3)/(sqrt(x^2+1)) ?

The integral of (9x^3)/(sqrt(x^2+1)) is 9(1/3 (x^2+1)^{3/2}-sqrt(x^2+1))+C

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integral of (9x^3)/(sqrt(x^2+1))

\int \frac{9 x ^{3}}{x ^{2} + 1} d x

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

Solution steps

\int \frac{9 x ^{3}}{x ^{2} + 1} d x

Take the constant out:

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

= 9 \cdot \int \frac{x ^{3}}{x ^{2} + 1} d x

Apply u-substitution

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

= 9 \cdot \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

= 9 (\int u^{2} d u - \int 1 d u)

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

\int 1 d u = u

= 9 (\frac{u ^{3}}{3} - u)

Substitute back

u = x^{2} + 1

= 9 (\frac{( x ^{2} + 1 ) ^{3}}{3} - x^{2} + 1)

Simplify

9 (\frac{( x ^{2} + 1 ) ^{3}}{3} - x^{2} + 1) : 9 (\frac{1}{3} (x^{2} + 1)^{\frac{3}{2}} - x^{2} + 1)

= 9 (\frac{1}{3} (x^{2} + 1)^{\frac{3}{2}} - x^{2} + 1)

Add a constant to the solution

= 9 (\frac{1}{3} (x^{2} + 1)^{\frac{3}{2}} - x^{2} + 1) + C

\int \frac{2 x - 3}{x ^{2} + 6 x + 4} d x

\frac{d}{d x} (13 e^{x})

f (x) = \frac{( x ^{2} )}{x - 5}, at x = - 3

0.001 t

t y^{^{'}} = 4 y

What is the integral of (9x^3)/(sqrt(x^2+1)) ?
The integral of (9x^3)/(sqrt(x^2+1)) is 9(1/3 (x^2+1)^{3/2}-sqrt(x^2+1))+C