What is the integral of x/(sqrt(9+8x^2-x^4)) ?

The integral of x/(sqrt(9+8x^2-x^4)) is 1/2 arcsin(1/5 (x^2-4))+C

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integral of x/(sqrt(9+8x^2-x^4))

\int \frac{x}{9 + 8 x ^{2} - x ^{4}} d x

\frac{1}{2} arcsin (\frac{1}{5} (x^{2} - 4)) + C

Solution steps

\int \frac{x}{9 + 8 x ^{2} - x ^{4}} d x

Apply u-substitution

: \int \frac{1}{2 9 + 8 u - u ^{2}} d u

= \int \frac{1}{2 9 + 8 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}{9 + 8 u - u ^{2}} d u

Complete the square

9 + 8 u - u^{2} : - (u - 4)^{2} + 25

= \frac{1}{2} \cdot \int \frac{1}{- ( u - 4 ) ^{2} + 25} d u

Apply u-substitution

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

= \frac{1}{2} \cdot \int \frac{1}{- v ^{2} + 25} 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}{5} (x^{2} - 4))

Simplify

= \frac{1}{2} arcsin (\frac{1}{5} (x^{2} - 4))

Add a constant to the solution

= \frac{1}{2} arcsin (\frac{1}{5} (x^{2} - 4)) + C

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What is the integral of x/(sqrt(9+8x^2-x^4)) ?
The integral of x/(sqrt(9+8x^2-x^4)) is 1/2 arcsin(1/5 (x^2-4))+C