What is the integral of (13x^3)/(sqrt(2+x^2)) ?

The integral of (13x^3)/(sqrt(2+x^2)) is 13(1/3 (2+x^2)^{3/2}-2sqrt(2+x^2))+C

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integral of (13x^3)/(sqrt(2+x^2))

\int \frac{13 x ^{3}}{2 + x ^{2}} d x

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

Solution steps

\int \frac{13 x ^{3}}{2 + x ^{2}} d x

Take the constant out:

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

= 13 \cdot \int \frac{x ^{3}}{2 + x ^{2}} d x

Apply u-substitution

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

= 13 \cdot \int u^{2} - 2 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

= 13 (\int u^{2} d u - \int 2 d u)

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

\int 2 d u = 2 u

= 13 (\frac{u ^{3}}{3} - 2 u)

Substitute back

u = 2 + x^{2}

= 13 (\frac{( 2 + x ^{2} ) ^{3}}{3} - 2 2 + x^{2})

Simplify

13 (\frac{( 2 + x ^{2} ) ^{3}}{3} - 2 2 + x^{2}) : 13 (\frac{1}{3} (2 + x^{2})^{\frac{3}{2}} - 2 2 + x^{2})

= 13 (\frac{1}{3} (2 + x^{2})^{\frac{3}{2}} - 2 2 + x^{2})

Add a constant to the solution

= 13 (\frac{1}{3} (2 + x^{2})^{\frac{3}{2}} - 2 2 + x^{2}) + C

sin ((π x)^{3})

x^{2} y^{^{'}} - 2 x y = 5 y^{4}, y (1) = \frac{1}{4}

\int_{1}^{2} \frac{y ^{3}}{2} + \frac{1}{2 y ^{3}} d y

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

\int_{0}^{\frac{π}{4}} cos^{2} (\frac{x}{3}) d x

What is the integral of (13x^3)/(sqrt(2+x^2)) ?
The integral of (13x^3)/(sqrt(2+x^2)) is 13(1/3 (2+x^2)^{3/2}-2sqrt(2+x^2))+C