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

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

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

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

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (e^{- 2 x} \cdot cos (x))

x \to 0 + lim (3 x^{(\frac{x}{2})})

\int \frac{2 x + 1}{( x + 1 ) ^{3} ( x ^{2} + 4 ) ^{2}} d x

\int x - 8 d x

x \to 4 lim (\frac{( x - 5 )}{x - 4})

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