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

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

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integral of (5x)/(sqrt(2-x^4))

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int 1 d v

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

Integral of a constant:

\int a d x = a x

= 5 \cdot \frac{1}{2} \cdot 1 \cdot v

Substitute back

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

Simplify

5 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (\frac{1}{2} x^{2}) : \frac{5}{2} arcsin (\frac{1}{2} x^{2})

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

Add a constant to the solution

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

\int \frac{ln ( t )}{t ^{3}} d t

\int sech^{2} (x) tanh (x) d x

x \to 3 - lim (\frac{2}{x ^{2} - 9})

e^{9 - x}

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