What is the integral of (4x)/(sqrt(11+10x-x^2)) ?

The integral of (4x)/(sqrt(11+10x-x^2)) is 4(-sqrt(-x^2+10x+11)+5arcsin(1/6 (x-5)))+C

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integral of (4x)/(sqrt(11+10x-x^2))

\int \frac{4 x}{11 + 10 x - x ^{2}} d x

4 (- - x^{2} + 10 x + 11 + 5 arcsin (\frac{1}{6} (x - 5))) + C

Solution steps

\int \frac{4 x}{11 + 10 x - x ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int \frac{x}{11 + 10 x - x ^{2}} d x

Complete the square

11 + 10 x - x^{2} : - (x - 5)^{2} + 36

= 4 \cdot \int \frac{x}{- ( x - 5 ) ^{2} + 36} d x

Apply u-substitution

: \int \frac{u + 5}{- u ^{2} + 36} d u

= 4 \cdot \int \frac{u + 5}{- u ^{2} + 36} d u

Apply Trigonometric Substitution

: \int 6 sin (v) + 5 d v

= 4 \cdot \int 6 sin (v) + 5 d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 (\int 6 sin (v) d v + \int 5 d v)

\int 6 sin (v) d v = - 6 cos (v)

\int 5 d v = 5 v

= 4 (- 6 cos (v) + 5 v)

Substitute back

= 4 (- 6 cos (arcsin (\frac{1}{6} (x - 5))) + 5 arcsin (\frac{1}{6} (x - 5)))

Simplify

4 (- 6 cos (arcsin (\frac{1}{6} (x - 5))) + 5 arcsin (\frac{1}{6} (x - 5))) : 4 (- - x^{2} + 10 x + 11 + 5 arcsin (\frac{1}{6} (x - 5)))

= 4 (- - x^{2} + 10 x + 11 + 5 arcsin (\frac{1}{6} (x - 5)))

Add a constant to the solution

= 4 (- - x^{2} + 10 x + 11 + 5 arcsin (\frac{1}{6} (x - 5))) + C

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What is the integral of (4x)/(sqrt(11+10x-x^2)) ?
The integral of (4x)/(sqrt(11+10x-x^2)) is 4(-sqrt(-x^2+10x+11)+5arcsin(1/6 (x-5)))+C