What is the integral of (sqrt(x)(x+1))/(4x^4) ?

The integral of (sqrt(x)(x+1))/(4x^4) is -(5x+3)/(30x^{5/2)}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (sqrt(x)(x+1))/(4x^4)

\int \frac{x ( x + 1 )}{4 x ^{4}} d x

- \frac{5 x + 3}{30 x ^{\frac{5}{2}}} + C

Solution steps

\int \frac{x ( x + 1 )}{4 x ^{4}} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{x ( x + 1 )}{x ^{4}} d x

Expand

\frac{x ( x + 1 )}{x ^{4}} : \frac{1}{x ^{\frac{5}{2}}} + \frac{1}{x ^{\frac{7}{2}}}

= \frac{1}{4} \cdot \int \frac{1}{x ^{\frac{5}{2}}} + \frac{1}{x ^{\frac{7}{2}}} d x

Apply the Sum Rule:

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

= \frac{1}{4} (\int \frac{1}{x ^{\frac{5}{2}}} d x + \int \frac{1}{x ^{\frac{7}{2}}} d x)

\int \frac{1}{x ^{\frac{5}{2}}} d x = - \frac{2}{3 x ^{\frac{3}{2}}}

\int \frac{1}{x ^{\frac{7}{2}}} d x = - \frac{2}{5 x ^{\frac{5}{2}}}

= \frac{1}{4} (- \frac{2}{3 x ^{\frac{3}{2}}} - \frac{2}{5 x ^{\frac{5}{2}}})

Simplify

\frac{1}{4} (- \frac{2}{3 x ^{\frac{3}{2}}} - \frac{2}{5 x ^{\frac{5}{2}}}) : - \frac{5 x + 3}{30 x ^{\frac{5}{2}}}

= - \frac{5 x + 3}{30 x ^{\frac{5}{2}}}

Add a constant to the solution

= - \frac{5 x + 3}{30 x ^{\frac{5}{2}}} + C

\frac{\partial}{\partial x} (y ln (x + y))

\int 16 e^{- 16 x} d x

(ln (x))^{l n (x)}

x \to 0 lim (\frac{10 sin ( 3 x )}{1 - cos ( 3 x )})

\frac{\partial}{\partial x} (e^{\frac{x ^{2}}{y ^{3}}})

What is the integral of (sqrt(x)(x+1))/(4x^4) ?
The integral of (sqrt(x)(x+1))/(4x^4) is -(5x+3)/(30x^{5/2)}+C