What is the integral of (10x^3-5x)/(sqrt(x^4-x^2+6)) ?

The integral of (10x^3-5x)/(sqrt(x^4-x^2+6)) is 5sqrt(x^4-x^2+6)+C

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integral of (10x^3-5x)/(sqrt(x^4-x^2+6))

\int \frac{10 x ^{3} - 5 x}{x ^{4} - x ^{2} + 6} d x

5 x^{4} - x^{2} + 6 + C

Solution steps

\int \frac{10 x ^{3} - 5 x}{x ^{4} - x ^{2} + 6} d x

Apply u-substitution

: \int \frac{5}{2 u} d u

= \int \frac{5}{2 u} d u

Take the constant out:

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

= \frac{5}{2} \cdot \int \frac{1}{u} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= \frac{5}{2} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = x^{4} - x^{2} + 6

= \frac{5}{2} \cdot 2 (x^{4} - x^{2} + 6)^{\frac{1}{2}}

Simplify

\frac{5}{2} \cdot 2 (x^{4} - x^{2} + 6)^{\frac{1}{2}} : 5 x^{4} - x^{2} + 6

= 5 x^{4} - x^{2} + 6

Add a constant to the solution

= 5 x^{4} - x^{2} + 6 + C

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What is the integral of (10x^3-5x)/(sqrt(x^4-x^2+6)) ?
The integral of (10x^3-5x)/(sqrt(x^4-x^2+6)) is 5sqrt(x^4-x^2+6)+C