What is the integral of 6/(sqrt(10x-x^2)) ?

The integral of 6/(sqrt(10x-x^2)) is 6arcsin(1/5 x-1)+C

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integral of 6/(sqrt(10x-x^2))

\int \frac{6}{10 x - x ^{2}} d x

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

Solution steps

\int \frac{6}{10 x - x ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int \frac{1}{10 x - x ^{2}} d x

Complete the square

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

= 6 \cdot \int \frac{1}{- ( x - 5 ) ^{2} + 25} d x

Apply u-substitution

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

= 6 \cdot \int \frac{1}{- u ^{2} + 25} d u

Apply Trigonometric Substitution

: \int 1 d v

= 6 \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 6 \cdot 1 \cdot v

Substitute back

= 6 \cdot 1 \cdot arcsin (\frac{1}{5} (x - 5))

Simplify

6 \cdot 1 \cdot arcsin (\frac{1}{5} (x - 5)) : 6 arcsin (\frac{1}{5} x - 1)

= 6 arcsin (\frac{1}{5} x - 1)

Add a constant to the solution

= 6 arcsin (\frac{1}{5} x - 1) + C

\frac{d}{d x} (x^{2} + \frac{128}{x})

y^{^{'''}} + 8 y^{^{''}} = - 6 x^{2} + 9 x + 2

ln (sec (2 x) + tan (2 x))

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

\int - 6 sin (6 x) d x

What is the integral of 6/(sqrt(10x-x^2)) ?
The integral of 6/(sqrt(10x-x^2)) is 6arcsin(1/5 x-1)+C