What is the integral of 1/(sqrt(-x^2+10x)) ?

The integral of 1/(sqrt(-x^2+10x)) is arcsin(1/5 x-1)+C

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integral of 1/(sqrt(-x^2+10x))

\int \frac{1}{- x ^{2} + 10 x} d x

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

Solution steps

\int \frac{1}{- x ^{2} + 10 x} d x

Complete the square

- x^{2} + 10 x : - (x - 5)^{2} + 25

= \int \frac{1}{- ( x - 5 ) ^{2} + 25} d x

Apply u-substitution

: \int \frac{1}{- u ^{2} + 25} d u

= \int \frac{1}{- u ^{2} + 25} 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 - 5))

Simplify

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

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

Add a constant to the solution

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

\int x^{7} \cdot sin (x^{4}) d x

f (x) = cos (x^{2})

n = 1 \sum \infty n (\frac{4}{3})^{n}

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

\frac{d y}{d t} = y + 2 t

What is the integral of 1/(sqrt(-x^2+10x)) ?
The integral of 1/(sqrt(-x^2+10x)) is arcsin(1/5 x-1)+C