What is the integral of (x^3)/(sqrt(x^2-25)) ?

The integral of (x^3)/(sqrt(x^2-25)) is 1/3 (x^2-25)^{3/2}+25sqrt(x^2-25)+C

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integral of (x^3)/(sqrt(x^2-25))

\int \frac{x ^{3}}{x ^{2} - 25} d x

\frac{1}{3} (x^{2} - 25)^{\frac{3}{2}} + 25 x^{2} - 25 + C

Solution steps

\int \frac{x ^{3}}{x ^{2} - 25} d x

Apply u-substitution

: \int u^{2} + 25 d u

= \int u^{2} + 25 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 25 d u

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

\int 25 d u = 25 u

= \frac{u ^{3}}{3} + 25 u

Substitute back

u = x^{2} - 25

= \frac{( x ^{2} - 25 ) ^{3}}{3} + 25 x^{2} - 25

Simplify

\frac{( x ^{2} - 25 ) ^{3}}{3} + 25 x^{2} - 25 : \frac{1}{3} (x^{2} - 25)^{\frac{3}{2}} + 25 x^{2} - 25

= \frac{1}{3} (x^{2} - 25)^{\frac{3}{2}} + 25 x^{2} - 25

Add a constant to the solution

= \frac{1}{3} (x^{2} - 25)^{\frac{3}{2}} + 25 x^{2} - 25 + C

x \to 0 lim (\frac{x ^{2} - 6}{x})

f (x) = \frac{x}{1 + x ^{2}}, (2, \frac{2}{5})

5 x + 11

\int \frac{2 x}{x ^{2} - 2 x + 5} d x

What is the integral of (x^3)/(sqrt(x^2-25)) ?
The integral of (x^3)/(sqrt(x^2-25)) is 1/3 (x^2-25)^{3/2}+25sqrt(x^2-25)+C