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

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

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integral of 1/(sqrt(3-x^2+4x))

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

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

Solution steps

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

Complete the square

3 - x^{2} + 4 x : - (x - 2)^{2} + 7

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

Apply u-substitution

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

Apply Trigonometric Substitution

= \int 1 d v

Integral of a constant:

\int a d x = a x

= 1 \cdot v

Substitute back

= 1 \cdot arcsin (\frac{1}{7} (x - 2))

Simplify

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

Add a constant to the solution

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

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