What is the integral of (1.2x)/(x^2sqrt(x^4-9)) ?

The integral of (1.2x)/(x^2sqrt(x^4-9)) is 0.2arctan(0.33333…sqrt(x^4-9))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (1.2x)/(x^2sqrt(x^4-9))

\int \frac{1.2 x}{x ^{2} x ^{4} - 9} d x

0.2 arctan (0.33333 \dots x^{4} - 9) + C

Solution steps

\int \frac{1.2 x}{x ^{2} x ^{4} - 9} d x

Take the constant out:

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

= 1.2 \cdot \int \frac{x}{x ^{2} x ^{4} - 9} d x

Cancel the common factor:

x

= 1.2 \cdot \int \frac{1}{x x ^{4} - 9} d x

Apply u-substitution

: \int \frac{1}{2 ( u ^{2} + 9 )} d u

= 1.2 \cdot \int \frac{1}{2 ( u ^{2} + 9 )} d u

Take the constant out:

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

= 1.2 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{2} + 9} d u

Apply Integral Substitution

= 1.2 \cdot \frac{1}{2} \cdot \int \frac{1}{3 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 1.2 \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= 1.2 \cdot \frac{1}{2} \cdot \frac{1}{3} arctan (v)

Substitute back

= 1.2 \cdot \frac{1}{2} \cdot \frac{1}{3} arctan (\frac{x ^{4} - 9}{3})

Simplify

1.2 \cdot \frac{1}{2} \cdot \frac{1}{3} arctan (\frac{x ^{4} - 9}{3}) : 0.2 arctan (0.33333 \dots x^{4} - 9)

= 0.2 arctan (0.33333 \dots x^{4} - 9)

Add a constant to the solution

= 0.2 arctan (0.33333 \dots x^{4} - 9) + C

(\frac{sin ( x )}{1 + x})^{^{'}}

y^{^{''}} + 2 y^{^{'}} + 2 y = 0, y (0) = - 2, y^{^{'}} (0) = - 4

(- x + 2) \cdot y^{^{'}} - y = e^{- 2 x}

\frac{d y}{d x} = a (y^{2} + 1)

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

What is the integral of (1.2x)/(x^2sqrt(x^4-9)) ?
The integral of (1.2x)/(x^2sqrt(x^4-9)) is 0.2arctan(0.33333…sqrt(x^4-9))+C