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

The integral of x/(sqrt(-5x^2+2x+1)) is 1/(5sqrt(5))(-sqrt(-25x^2+10x+5)+arcsin((5x-1)/(sqrt(6))))+C

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integral of x/(sqrt(-5x^2+2x+1))

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

\frac{1}{5 5} (- - 25 x^{2} + 10 x + 5 + arcsin (\frac{5 x - 1}{6})) + C

Solution steps

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

Complete the square

- 5 x^{2} + 2 x + 1 : - 5 (x - \frac{1}{5})^{2} + \frac{6}{5}

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

Apply u-substitution

: \int \frac{5 u + 1}{5 - 25 u ^{2} + 6} d u

= \int \frac{5 u + 1}{5 - 25 u ^{2} + 6} d u

Take the constant out:

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

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

Expand

\frac{5 u + 1}{- 25 u ^{2} + 6} : \frac{5 u}{- 25 u ^{2} + 6} + \frac{1}{- 25 u ^{2} + 6}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{5 u}{- 25 u ^{2} + 6} d u = - \frac{1}{5} - 25 u^{2} + 6

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

= \frac{1}{5} (- \frac{1}{5} - 25 u^{2} + 6 + \frac{1}{5} arcsin (\frac{5}{6} u))

Substitute back

u = x - \frac{1}{5}

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

Simplify

\frac{1}{5} - \frac{1}{5} - 25 (x - \frac{1}{5})^{2} + 6 + \frac{1}{5} arcsin (\frac{5}{6} (x - \frac{1}{5})) : \frac{1}{5 5} (- - 25 x^{2} + 10 x + 5 + arcsin (\frac{5 x - 1}{6}))

= \frac{1}{5 5} (- - 25 x^{2} + 10 x + 5 + arcsin (\frac{5 x - 1}{6}))

Add a constant to the solution

= \frac{1}{5 5} (- - 25 x^{2} + 10 x + 5 + arcsin (\frac{5 x - 1}{6})) + C

\frac{d}{d x} (8 (x + 4)^{\frac{3}{2}})

x \to - 2 lim (\frac{x + 1}{1 - x ^{2}})

\frac{d}{d x} (r (x) cos (x))

\int y \cdot cos (x y) d x

n = 0 \sum \infty 1

What is the integral of x/(sqrt(-5x^2+2x+1)) ?
The integral of x/(sqrt(-5x^2+2x+1)) is 1/(5sqrt(5))(-sqrt(-25x^2+10x+5)+arcsin((5x-1)/(sqrt(6))))+C