What is the integral of (7x-2)/(sqrt(5-4x-x^2)) ?

The integral of (7x-2)/(sqrt(5-4x-x^2)) is -7sqrt(-x^2-4x+5)-16arcsin(1/3 (x+2))+C

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integral of (7x-2)/(sqrt(5-4x-x^2))

\int \frac{7 x - 2}{5 - 4 x - x ^{2}} d x

- 7 - x^{2} - 4 x + 5 - 16 arcsin (\frac{1}{3} (x + 2)) + C

Solution steps

\int \frac{7 x - 2}{5 - 4 x - x ^{2}} d x

Expand

\frac{7 x - 2}{5 - 4 x - x ^{2}} : \frac{7 x}{5 - 4 x - x ^{2}} - \frac{2}{5 - 4 x - x ^{2}}

= \int \frac{7 x}{5 - 4 x - x ^{2}} - \frac{2}{5 - 4 x - 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{7 x}{5 - 4 x - x ^{2}} d x - \int \frac{2}{5 - 4 x - x ^{2}} d x

\int \frac{7 x}{5 - 4 x - x ^{2}} d x = 7 (- - x^{2} - 4 x + 5 - 2 arcsin (\frac{1}{3} (x + 2)))

\int \frac{2}{5 - 4 x - x ^{2}} d x = 2 arcsin (\frac{1}{3} (x + 2))

= 7 (- - x^{2} - 4 x + 5 - 2 arcsin (\frac{1}{3} (x + 2))) - 2 arcsin (\frac{1}{3} (x + 2))

Simplify

7 (- - x^{2} - 4 x + 5 - 2 arcsin (\frac{1}{3} (x + 2))) - 2 arcsin (\frac{1}{3} (x + 2)) : - 7 - x^{2} - 4 x + 5 - 16 arcsin (\frac{1}{3} (x + 2))

= - 7 - x^{2} - 4 x + 5 - 16 arcsin (\frac{1}{3} (x + 2))

Add a constant to the solution

= - 7 - x^{2} - 4 x + 5 - 16 arcsin (\frac{1}{3} (x + 2)) + C

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What is the integral of (7x-2)/(sqrt(5-4x-x^2)) ?
The integral of (7x-2)/(sqrt(5-4x-x^2)) is -7sqrt(-x^2-4x+5)-16arcsin(1/3 (x+2))+C