What is the integral of 1/(sqrt(9+16x-4x^2)) ?

The integral of 1/(sqrt(9+16x-4x^2)) is 1/2 arcsin(2/5 (x-2))+C

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integral of 1/(sqrt(9+16x-4x^2))

\int \frac{1}{9 + 16 x - 4 x ^{2}} d x

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

Solution steps

\int \frac{1}{9 + 16 x - 4 x ^{2}} d x

Complete the square

9 + 16 x - 4 x^{2} : - 4 (x - 2)^{2} + 25

= \int \frac{1}{- 4 ( x - 2 ) ^{2} + 25} d x

Apply u-substitution

: \int \frac{1}{- 4 u ^{2} + 25} d u

= \int \frac{1}{- 4 u ^{2} + 25} d u

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Integral of a constant:

\int a d x = a x

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

Substitute back

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

Simplify

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

Add a constant to the solution

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

\int \frac{x ^{5} - 2 x}{x ^{3}} d x

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\int 45 (1 + x^{2}) (7 + 15 x + 5 x^{3})^{10} d x

\int \frac{x ^{3} - 4 x + 6}{x + 7} d x

What is the integral of 1/(sqrt(9+16x-4x^2)) ?
The integral of 1/(sqrt(9+16x-4x^2)) is 1/2 arcsin(2/5 (x-2))+C