What is the integral of (7x^3)/(sqrt(4+x^2)) ?

The integral of (7x^3)/(sqrt(4+x^2)) is 7(1/3 (4+x^2)^{3/2}-4sqrt(4+x^2))+C

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integral of (7x^3)/(sqrt(4+x^2))

\int \frac{7 x ^{3}}{4 + x ^{2}} d x

7 (\frac{1}{3} (4 + x^{2})^{\frac{3}{2}} - 4 4 + x^{2}) + C

Solution steps

\int \frac{7 x ^{3}}{4 + x ^{2}} d x

Take the constant out:

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

= 7 \cdot \int \frac{x ^{3}}{4 + x ^{2}} d x

Apply u-substitution

: \int u^{2} - 4 d u

= 7 \cdot \int u^{2} - 4 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

= 7 (\int u^{2} d u - \int 4 d u)

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

\int 4 d u = 4 u

= 7 (\frac{u ^{3}}{3} - 4 u)

Substitute back

u = 4 + x^{2}

= 7 (\frac{( 4 + x ^{2} ) ^{3}}{3} - 4 4 + x^{2})

Simplify

7 (\frac{( 4 + x ^{2} ) ^{3}}{3} - 4 4 + x^{2}) : 7 (\frac{1}{3} (4 + x^{2})^{\frac{3}{2}} - 4 4 + x^{2})

= 7 (\frac{1}{3} (4 + x^{2})^{\frac{3}{2}} - 4 4 + x^{2})

Add a constant to the solution

= 7 (\frac{1}{3} (4 + x^{2})^{\frac{3}{2}} - 4 4 + x^{2}) + C

y = - 2 x^{3}, (1, - 2)

\int (3 x^{2}) d x

\frac{\partial}{\partial x} (sin^{2} (3 x y^{2} - y))

\frac{d y}{d x}, y = 1 1^{x}

\frac{d}{d x} ((x^{2} - 3) e^{x})

What is the integral of (7x^3)/(sqrt(4+x^2)) ?
The integral of (7x^3)/(sqrt(4+x^2)) is 7(1/3 (4+x^2)^{3/2}-4sqrt(4+x^2))+C