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

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

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

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

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

Solution steps

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

Complete the square

1 + 4 x - x^{2} : - (x - 2)^{2} + 5

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

Apply u-substitution

: \int \frac{1}{- u ^{2} + 5} d u

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

Apply Trigonometric Substitution

: \int 1 d v

= \int 1 d v

Integral of a constant:

\int a d x = a x

= 1 \cdot v

Substitute back

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

Simplify

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

Add a constant to the solution

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

\int sec (\frac{1}{2}) y d y

\int (1 - x^{2})^{3} d x

\int \frac{x ^{3}}{( x + 1 )} d x

\int 9 cos^{2} (3 x) d x

\int \frac{1}{5 + cos ( x )} d x

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