What is the integral of 1/(sqrt(-x^2+8x-15)) ?

The integral of 1/(sqrt(-x^2+8x-15)) is arcsin(x-4)+C

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integral of 1/(sqrt(-x^2+8x-15))

\int \frac{1}{- x ^{2} + 8 x - 15} d x

arcsin (x - 4) + C

Solution steps

\int \frac{1}{- x ^{2} + 8 x - 15} d x

Complete the square

- x^{2} + 8 x - 15 : - (x - 4)^{2} + 1

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{- u ^{2} + 1} d u = arcsin (u)

= arcsin (u)

Substitute back

u = x - 4

= arcsin (x - 4)

Add a constant to the solution

= arcsin (x - 4) + C

\int \frac{e ^{3 x} + 3 e ^{x} + 2}{e ^{x}} d x

f (t) = - \frac{1}{t ^{2}}

\int 2 arctan (3 x) d x

\int - x e^{- l n (x)} d x

f (x) = 2 x^{3} + 3 x^{2}, at x = 1

What is the integral of 1/(sqrt(-x^2+8x-15)) ?
The integral of 1/(sqrt(-x^2+8x-15)) is arcsin(x-4)+C