What is the integral of (x^3)/(sqrt(x^2+49)) ?

The integral of (x^3)/(sqrt(x^2+49)) is 1/3 (x^2+49)^{3/2}-49sqrt(x^2+49)+C

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integral of (x^3)/(sqrt(x^2+49))

\int \frac{x ^{3}}{x ^{2} + 49} d x

\frac{1}{3} (x^{2} + 49)^{\frac{3}{2}} - 49 x^{2} + 49 + C

Solution steps

\int \frac{x ^{3}}{x ^{2} + 49} d x

Apply u-substitution

= \int u^{2} - 49 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 49 d u

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

\int 49 d u = 49 u

= \frac{u ^{3}}{3} - 49 u

Substitute back

u = x^{2} + 49

= \frac{( x ^{2} + 49 ) ^{3}}{3} - 49 x^{2} + 49

Simplify

\frac{( x ^{2} + 49 ) ^{3}}{3} - 49 x^{2} + 49 : \frac{1}{3} (x^{2} + 49)^{\frac{3}{2}} - 49 x^{2} + 49

= \frac{1}{3} (x^{2} + 49)^{\frac{3}{2}} - 49 x^{2} + 49

Add a constant to the solution

= \frac{1}{3} (x^{2} + 49)^{\frac{3}{2}} - 49 x^{2} + 49 + C

\int (\frac{1}{2} x) d x

t a y l or e^{- h} + e^{h}

d er i v a t i v e 20 (1 + e^{10 - t})^{- 1}

y^{^{'}} = 8 x y

\frac{\partial}{\partial x} (x^{4} + y^{2} - 4)

What is the integral of (x^3)/(sqrt(x^2+49)) ?
The integral of (x^3)/(sqrt(x^2+49)) is 1/3 (x^2+49)^{3/2}-49sqrt(x^2+49)+C