What is the integral of ((x^2))/(sqrt(13-4x+x^2)) ?

The integral of ((x^2))/(sqrt(13-4x+x^2)) is

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integral of ((x^2))/(sqrt(13-4x+x^2))

\int \frac{( x ^{2} )}{13 - 4 x + x ^{2}} d x

9 (- \frac{1}{2} ln (\frac{x - 2 + x ^{2} - 4 x + 13}{3}) + \frac{1}{18} (x - 2) x^{2} - 4 x + 13) + 4 x^{2} - 4 x + 13 + 4 ln (\frac{1}{3} x - 2 + x^{2} - 4 x + 13) + C

Solution steps

\int \frac{x ^{2}}{13 - 4 x + x ^{2}} d x

Complete the square

13 - 4 x + x^{2} : (x - 2)^{2} + 9

= \int \frac{x ^{2}}{( x - 2 ) ^{2} + 9} d x

Apply u-substitution

: \int \frac{( u + 2 ) ^{2}}{u ^{2} + 9} d u

= \int \frac{( u + 2 ) ^{2}}{u ^{2} + 9} d u

Expand

\frac{( u + 2 ) ^{2}}{u ^{2} + 9} : \frac{u ^{2}}{u ^{2} + 9} + \frac{4 u}{u ^{2} + 9} + \frac{4}{u ^{2} + 9}

= \int \frac{u ^{2}}{u ^{2} + 9} + \frac{4 u}{u ^{2} + 9} + \frac{4}{u ^{2} + 9} 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 \frac{u ^{2}}{u ^{2} + 9} d u + \int \frac{4 u}{u ^{2} + 9} d u + \int \frac{4}{u ^{2} + 9} d u

\int \frac{u ^{2}}{u ^{2} + 9} d u = 9 (- \frac{1}{2} ln (\frac{u + 9 + u ^{2}}{3}) + \frac{1}{18} u 9 + u^{2})

\int \frac{4 u}{u ^{2} + 9} d u = 4 u^{2} + 9

\int \frac{4}{u ^{2} + 9} d u = 4 ln (\frac{1}{3} u + 9 + u^{2})

= 9 (- \frac{1}{2} ln (\frac{u + 9 + u ^{2}}{3}) + \frac{1}{18} u 9 + u^{2}) + 4 u^{2} + 9 + 4 ln (\frac{1}{3} u + 9 + u^{2})

Substitute back

u = x - 2

= 9 - \frac{1}{2} ln \frac{x - 2 + 9 + ( x - 2 ) ^{2}}{3} + \frac{1}{18} (x - 2) 9 + (x - 2)^{2} + 4 (x - 2)^{2} + 9 + 4 ln (\frac{1}{3} x - 2 + 9 + (x - 2)^{2})

Simplify

9 - \frac{1}{2} ln \frac{x - 2 + 9 + ( x - 2 ) ^{2}}{3} + \frac{1}{18} (x - 2) 9 + (x - 2)^{2} + 4 (x - 2)^{2} + 9 + 4 ln (\frac{1}{3} x - 2 + 9 + (x - 2)^{2}) : 9 (- \frac{1}{2} ln (\frac{x - 2 + x ^{2} - 4 x + 13}{3}) + \frac{1}{18} (x - 2) x^{2} - 4 x + 13) + 4 x^{2} - 4 x + 13 + 4 ln (\frac{1}{3} x - 2 + x^{2} - 4 x + 13)

= 9 (- \frac{1}{2} ln (\frac{x - 2 + x ^{2} - 4 x + 13}{3}) + \frac{1}{18} (x - 2) x^{2} - 4 x + 13) + 4 x^{2} - 4 x + 13 + 4 ln (\frac{1}{3} x - 2 + x^{2} - 4 x + 13)

Add a constant to the solution

= 9 (- \frac{1}{2} ln (\frac{x - 2 + x ^{2} - 4 x + 13}{3}) + \frac{1}{18} (x - 2) x^{2} - 4 x + 13) + 4 x^{2} - 4 x + 13 + 4 ln (\frac{1}{3} x - 2 + x^{2} - 4 x + 13) + C

\frac{d}{d x} (\frac{x + 6}{x ^{3} + x - 1})

y^{^{''}} + 4 y^{^{'}} + 3 y = t e^{- t}

x \to 6 lim (\frac{3 x}{x - 6})

\int r^{2} e^{r} d r

\frac{d}{d x} (cos (arctan (x)))

What is the integral of ((x^2))/(sqrt(13-4x+x^2)) ?
The integral of ((x^2))/(sqrt(13-4x+x^2)) is