What is the integral of (4x)/(sqrt(16-x^4)) ?

The integral of (4x)/(sqrt(16-x^4)) is 2arcsin(1/4 x^2)+C

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integral of (4x)/(sqrt(16-x^4))

\int \frac{4 x}{16 - x ^{4}} d x

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

Solution steps

\int \frac{4 x}{16 - x ^{4}} d x

Take the constant out:

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

= 4 \cdot \int \frac{x}{16 - x ^{4}} d x

Apply u-substitution

: \int \frac{1}{2 16 - u ^{2}} d u

= 4 \cdot \int \frac{1}{2 16 - u ^{2}} d u

Take the constant out:

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

= 4 \cdot \frac{1}{2} \cdot \int \frac{1}{16 - u ^{2}} d u

Apply Trigonometric Substitution

: \int 1 d v

= 4 \cdot \frac{1}{2} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 4 \cdot \frac{1}{2} \cdot 1 \cdot v

Substitute back

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

Simplify

4 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (\frac{1}{4} x^{2}) : 2 arcsin (\frac{1}{4} x^{2})

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

Add a constant to the solution

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

y^{^{'}} = 8 y^{2} cos (t) + 7 y^{2} sin (4 t)

x^{3} - 2 x + 2

\int sin (4 x) cos (17 x) d x

n = 0 \sum \infty (\frac{2}{11})^{n}

- 6 x^{2} + 5 x

What is the integral of (4x)/(sqrt(16-x^4)) ?
The integral of (4x)/(sqrt(16-x^4)) is 2arcsin(1/4 x^2)+C