What is the integral of (x+(arccos(3x))^2)/(sqrt(1-9x^2)) ?

The integral of (x+(arccos(3x))^2)/(sqrt(1-9x^2)) is -1/9 sqrt(1-9x^2)-1/9 arccos^3(3x)+C

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integral of (x+(arccos(3x))^2)/(sqrt(1-9x^2))

\int \frac{x + ( arccos ( 3 x ) ) ^{2}}{1 - 9 x ^{2}} d x

- \frac{1}{9} 1 - 9 x^{2} - \frac{1}{9} arccos^{3} (3 x) + C

Solution steps

\int \frac{x + ( arccos ( 3 x ) ) ^{2}}{1 - 9 x ^{2}} d x

\frac{x + ( arccos ( 3 x ) ) ^{2}}{1 - 9 x ^{2}} = (\frac{x}{1 - 9 x ^{2}} + \frac{( arccos ( 3 x ) ) ^{2}}{1 - 9 x ^{2}})

= \int \frac{x}{1 - 9 x ^{2}} + \frac{( arccos ( 3 x ) ) ^{2}}{1 - 9 x ^{2}} d x

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{x}{1 - 9 x ^{2}} d x + \int \frac{( arccos ( 3 x ) ) ^{2}}{1 - 9 x ^{2}} d x

\int \frac{x}{1 - 9 x ^{2}} d x = - \frac{1}{9} 1 - 9 x^{2}

\int \frac{( arccos ( 3 x ) ) ^{2}}{1 - 9 x ^{2}} d x = - \frac{1}{9} arccos^{3} (3 x)

= - \frac{1}{9} 1 - 9 x^{2} - \frac{1}{9} arccos^{3} (3 x)

Add a constant to the solution

= - \frac{1}{9} 1 - 9 x^{2} - \frac{1}{9} arccos^{3} (3 x) + C

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What is the integral of (x+(arccos(3x))^2)/(sqrt(1-9x^2)) ?
The integral of (x+(arccos(3x))^2)/(sqrt(1-9x^2)) is -1/9 sqrt(1-9x^2)-1/9 arccos^3(3x)+C