What is the integral of x/(sqrt(8-2x^2-x^4)) ?

The integral of x/(sqrt(8-2x^2-x^4)) is 1/2 arcsin(1/3 (x^2+1))+C

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integral of x/(sqrt(8-2x^2-x^4))

\int \frac{x}{8 - 2 x ^{2} - x ^{4}} d x

\frac{1}{2} arcsin (\frac{1}{3} (x^{2} + 1)) + C

Solution steps

\int \frac{x}{8 - 2 x ^{2} - x ^{4}} d x

Apply u-substitution

: \int \frac{1}{2 8 - 2 u - u ^{2}} d u

= \int \frac{1}{2 8 - 2 u - u ^{2}} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{8 - 2 u - u ^{2}} d u

Complete the square

8 - 2 u - u^{2} : - (u + 1)^{2} + 9

= \frac{1}{2} \cdot \int \frac{1}{- ( u + 1 ) ^{2} + 9} d u

Apply u-substitution

: \int \frac{1}{- v ^{2} + 9} d v

= \frac{1}{2} \cdot \int \frac{1}{- v ^{2} + 9} d v

Apply Trigonometric Substitution

: \int 1 d w

= \frac{1}{2} \cdot \int 1 d w

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot 1 \cdot w

Substitute back

= \frac{1}{2} \cdot 1 \cdot arcsin (\frac{1}{3} (x^{2} + 1))

Simplify

= \frac{1}{2} arcsin (\frac{1}{3} (x^{2} + 1))

Add a constant to the solution

= \frac{1}{2} arcsin (\frac{1}{3} (x^{2} + 1)) + C

\int sin (7 x) d x

y^{^{''}} + 4 y = 4 t

f (x) = (7^{- 3 x})

What is the integral of x/(sqrt(8-2x^2-x^4)) ?
The integral of x/(sqrt(8-2x^2-x^4)) is 1/2 arcsin(1/3 (x^2+1))+C