What is the integral of (5x^3-4x)/(sqrt(1-x^4)) ?

The integral of (5x^3-4x)/(sqrt(1-x^4)) is -5/2 sqrt(1-x^4)-2arcsin(x^2)+C

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integral of (5x^3-4x)/(sqrt(1-x^4))

\int \frac{5 x ^{3} - 4 x}{1 - x ^{4}} d x

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

Solution steps

\int \frac{5 x ^{3} - 4 x}{1 - x ^{4}} d x

Expand

\frac{5 x ^{3} - 4 x}{1 - x ^{4}} : \frac{5 x ^{3}}{1 - x ^{4}} - \frac{4 x}{1 - x ^{4}}

= \int \frac{5 x ^{3}}{1 - x ^{4}} - \frac{4 x}{1 - x ^{4}} d x

Apply the Sum Rule:

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

= \int \frac{5 x ^{3}}{1 - x ^{4}} d x - \int \frac{4 x}{1 - x ^{4}} d x

\int \frac{5 x ^{3}}{1 - x ^{4}} d x = - \frac{5}{2} 1 - x^{4}

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

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

Add a constant to the solution

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

\int_{0}^{1} \frac{x - 7}{x ^{2} - 5 x + 6} d x

y^{^{'}} + 6 y = e^{4 t}

\frac{d}{d x} (sec^{5} (x^{2} - 1))

3 x^{2} + 1

f (x) = x^{\frac{1}{9}}

What is the integral of (5x^3-4x)/(sqrt(1-x^4)) ?
The integral of (5x^3-4x)/(sqrt(1-x^4)) is -5/2 sqrt(1-x^4)-2arcsin(x^2)+C