What is the integral of (6-72x)/(sqrt(1-36x^2)) ?

The integral of (6-72x)/(sqrt(1-36x^2)) is arcsin(6x)+2sqrt(1-36x^2)+C

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integral of (6-72x)/(sqrt(1-36x^2))

\int \frac{6 - 72 x}{1 - 36 x ^{2}} d x

arcsin (6 x) + 2 1 - 36 x^{2} + C

Solution steps

\int \frac{6 - 72 x}{1 - 36 x ^{2}} d x

Expand

\frac{6 - 72 x}{1 - 36 x ^{2}} : \frac{6}{1 - 36 x ^{2}} - \frac{72 x}{1 - 36 x ^{2}}

= \int \frac{6}{1 - 36 x ^{2}} - \frac{72 x}{1 - 36 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{6}{1 - 36 x ^{2}} d x - \int \frac{72 x}{1 - 36 x ^{2}} d x

\int \frac{6}{1 - 36 x ^{2}} d x = arcsin (6 x)

\int \frac{72 x}{1 - 36 x ^{2}} d x = - 2 1 - 36 x^{2}

= arcsin (6 x) - (- 2 1 - 36 x^{2})

Simplify

= arcsin (6 x) + 2 1 - 36 x^{2}

Add a constant to the solution

= arcsin (6 x) + 2 1 - 36 x^{2} + C

\frac{d}{d x} (\frac{x}{x + 2})

\frac{d}{d x} (\frac{x + 4}{x - 5})

23 x

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

\int_{1}^{2} \frac{1}{x ^{8}} d x

What is the integral of (6-72x)/(sqrt(1-36x^2)) ?
The integral of (6-72x)/(sqrt(1-36x^2)) is arcsin(6x)+2sqrt(1-36x^2)+C