What is the integral of 5/(sqrt(6+8x-4x^2)) ?

The integral of 5/(sqrt(6+8x-4x^2)) is 5/2 arcsin((sqrt(2))/(sqrt(5))(x-1))+C

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integral of 5/(sqrt(6+8x-4x^2))

\int \frac{5}{6 + 8 x - 4 x ^{2}} d x

\frac{5}{2} arcsin (\frac{2}{5} (x - 1)) + C

Solution steps

\int \frac{5}{6 + 8 x - 4 x ^{2}} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{6 + 8 x - 4 x ^{2}} d x

Complete the square

6 + 8 x - 4 x^{2} : - 4 (x - 1)^{2} + 10

= 5 \cdot \int \frac{1}{- 4 ( x - 1 ) ^{2} + 10} d x

Apply u-substitution

: \int \frac{1}{- 4 u ^{2} + 10} d u

= 5 \cdot \int \frac{1}{- 4 u ^{2} + 10} d u

Apply Trigonometric Substitution

: \int \frac{1}{2} d v

= 5 \cdot \int \frac{1}{2} d v

Take the constant out:

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

= 5 \cdot \frac{1}{2} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 5 \cdot \frac{1}{2} \cdot 1 \cdot v

Substitute back

= 5 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (\frac{2}{10} (x - 1))

Simplify

5 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (\frac{2}{10} (x - 1)) : \frac{5}{2} arcsin (\frac{2}{5} (x - 1))

= \frac{5}{2} arcsin (\frac{2}{5} (x - 1))

Add a constant to the solution

= \frac{5}{2} arcsin (\frac{2}{5} (x - 1)) + C

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What is the integral of 5/(sqrt(6+8x-4x^2)) ?
The integral of 5/(sqrt(6+8x-4x^2)) is 5/2 arcsin((sqrt(2))/(sqrt(5))(x-1))+C