What is the integral of 5/(x(x^2+6)^2) ?

The integral of 5/(x(x^2+6)^2) is 5/2 (1/18 ln(x)-1/36 ln|x^2+6|+1/(6(x^2+6)))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 5/(x(x^2+6)^2)

\int \frac{5}{x ( x ^{2} + 6 ) ^{2}} d x

\frac{5}{2} (\frac{1}{18} ln (x) - \frac{1}{36} ln x^{2} + 6 + \frac{1}{6 ( x ^{2} + 6 )}) + C

Solution steps

\int \frac{5}{x ( x ^{2} + 6 ) ^{2}} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{x ( x ^{2} + 6 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{2 ( u - 6 ) u ^{2}} d u

= 5 \cdot \int \frac{1}{2 ( u - 6 ) 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}{( u - 6 ) u ^{2}} d u

Take the partial fraction of

\frac{1}{( u - 6 ) u ^{2}} : \frac{1}{36 ( u - 6 )} - \frac{1}{36 u} - \frac{1}{6 u ^{2}}

= 5 \cdot \frac{1}{2} \cdot \int \frac{1}{36 ( u - 6 )} - \frac{1}{36 u} - \frac{1}{6 u ^{2}} d u

Apply the Sum Rule:

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

= 5 \cdot \frac{1}{2} (\int \frac{1}{36 ( u - 6 )} d u - \int \frac{1}{36 u} d u - \int \frac{1}{6 u ^{2}} d u)

\int \frac{1}{36 ( u - 6 )} d u = \frac{1}{36} ln ∣ u - 6 ∣

\int \frac{1}{36 u} d u = \frac{1}{36} ln ∣ u ∣

\int \frac{1}{6 u ^{2}} d u = - \frac{1}{6 u}

= 5 \cdot \frac{1}{2} (\frac{1}{36} ln ∣ u - 6 ∣ - \frac{1}{36} ln ∣ u ∣ - (- \frac{1}{6 u}))

Substitute back

u = x^{2} + 6

= 5 \cdot \frac{1}{2} (\frac{1}{36} ln x^{2} + 6 - 6 - \frac{1}{36} ln x^{2} + 6 - (- \frac{1}{6 ( x ^{2} + 6 )}))

Simplify

5 \cdot \frac{1}{2} (\frac{1}{36} ln x^{2} + 6 - 6 - \frac{1}{36} ln x^{2} + 6 - (- \frac{1}{6 ( x ^{2} + 6 )})) : \frac{5}{2} (\frac{1}{18} ln (x) - \frac{1}{36} ln x^{2} + 6 + \frac{1}{6 ( x ^{2} + 6 )})

= \frac{5}{2} (\frac{1}{18} ln (x) - \frac{1}{36} ln x^{2} + 6 + \frac{1}{6 ( x ^{2} + 6 )})

Add a constant to the solution

= \frac{5}{2} (\frac{1}{18} ln (x) - \frac{1}{36} ln x^{2} + 6 + \frac{1}{6 ( x ^{2} + 6 )}) + C

\int \frac{( y - 3 )}{y ^{2} - 6 y + 1} d y

\int cos^{6} (\frac{x}{2}) sin^{4} (\frac{x}{2}) d x

\int \frac{sin ( θ )}{cos ^{2} ( θ ) + cos ( θ ) - 2} d θ

\int 32 arctan (x) d x

\int x^{2} 3 x^{3} - 2 d x

What is the integral of 5/(x(x^2+6)^2) ?
The integral of 5/(x(x^2+6)^2) is 5/2 (1/18 ln(x)-1/36 ln|x^2+6|+1/(6(x^2+6)))+C