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

(\frac{x ^{2}}{e ^{2 x}})^{6}

- 7 x + \frac{3}{x ^{6}}

\frac{d y}{d x} = e^{3 x - 2 y}

\frac{d}{d x} (f (x) - 4 x + 1)

n = 1 \sum \infty (\frac{10}{11})^{n}

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