What is the integral of 1/(sqrt(-x^2+6x+7)) ?

The integral of 1/(sqrt(-x^2+6x+7)) is arcsin(1/4 (x-3))+C

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integral of 1/(sqrt(-x^2+6x+7))

\int \frac{1}{- x ^{2} + 6 x + 7} d x

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

Solution steps

\int \frac{1}{- x ^{2} + 6 x + 7} d x

Complete the square

- x^{2} + 6 x + 7 : - (x - 3)^{2} + 16

= \int \frac{1}{- ( x - 3 ) ^{2} + 16} d x

Apply u-substitution

: \int \frac{1}{- u ^{2} + 16} d u

= \int \frac{1}{- u ^{2} + 16} 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}{4} (x - 3))

Simplify

= arcsin (\frac{1}{4} (x - 3))

Add a constant to the solution

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

\int 5 x^{- 3} + 4 x^{- 2} - 3 x^{- 1} + 4 d x

\int_{0}^{1} 1 + \frac{9 x}{4} d x

cos (t) y^{^{'}} + sin (t) y = 2 cos^{3} (t) sin (t) - 1

\frac{c}{x}

6 x^{2} ln (6 x)

What is the integral of 1/(sqrt(-x^2+6x+7)) ?
The integral of 1/(sqrt(-x^2+6x+7)) is arcsin(1/4 (x-3))+C