What is the integral of (x+3)/(sqrt(3+4x-4x^2)) ?

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

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integral of (x+3)/(sqrt(3+4x-4x^2))

\int \frac{x + 3}{3 + 4 x - 4 x ^{2}} d x

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

Solution steps

\int \frac{x + 3}{3 + 4 x - 4 x ^{2}} d x

Complete the square

3 + 4 x - 4 x^{2} : - 4 (x - \frac{1}{2})^{2} + 4

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

Apply u-substitution

= \int \frac{2 u + 7}{4 - u ^{2} + 1} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{2 u + 7}{- u ^{2} + 1} d u

Expand

\frac{2 u + 7}{- u ^{2} + 1} : \frac{2 u}{- u ^{2} + 1} + \frac{7}{- u ^{2} + 1}

= \frac{1}{4} \cdot \int \frac{2 u}{- u ^{2} + 1} + \frac{7}{- u ^{2} + 1} 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

= \frac{1}{4} (\int \frac{2 u}{- u ^{2} + 1} d u + \int \frac{7}{- u ^{2} + 1} d u)

\int \frac{2 u}{- u ^{2} + 1} d u = - 2 - u^{2} + 1

\int \frac{7}{- u ^{2} + 1} d u = 7 arcsin (u)

= \frac{1}{4} (- 2 - u^{2} + 1 + 7 arcsin (u))

Substitute back

u = x - \frac{1}{2}

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

Simplify

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

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

Add a constant to the solution

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

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