What is the integral of (205x)/(sqrt(25-x^4)) ?

The integral of (205x)/(sqrt(25-x^4)) is 205/2 arcsin(1/5 x^2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (205x)/(sqrt(25-x^4))

\int \frac{205 x}{25 - x ^{4}} d x

\frac{205}{2} arcsin (\frac{1}{5} x^{2}) + C

Solution steps

\int \frac{205 x}{25 - x ^{4}} d x

Take the constant out:

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

= 205 \cdot \int \frac{x}{25 - x ^{4}} d x

Apply u-substitution

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

= 205 \cdot \int \frac{1}{2 25 - u ^{2}} d u

Take the constant out:

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

= 205 \cdot \frac{1}{2} \cdot \int \frac{1}{25 - u ^{2}} d u

Apply Trigonometric Substitution

: \int 1 d v

= 205 \cdot \frac{1}{2} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 205 \cdot \frac{1}{2} \cdot 1 \cdot v

Substitute back

= 205 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (\frac{1}{5} x^{2})

Simplify

205 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (\frac{1}{5} x^{2}) : \frac{205}{2} arcsin (\frac{1}{5} x^{2})

= \frac{205}{2} arcsin (\frac{1}{5} x^{2})

Add a constant to the solution

= \frac{205}{2} arcsin (\frac{1}{5} x^{2}) + C

\frac{\partial}{\partial x} (x^{- 3} y^{- 4})

x, 0

(x + 4 y) d x + 2 x d y = 0

y = \frac{4}{1 + x ^{2}}

What is the integral of (205x)/(sqrt(25-x^4)) ?
The integral of (205x)/(sqrt(25-x^4)) is 205/2 arcsin(1/5 x^2)+C