What is the integral of (x^4+1)/(2x^2) ?

The integral of (x^4+1)/(2x^2) is 1/2 ((x^3)/3-1/x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^4+1)/(2x^2)

\int \frac{x ^{4} + 1}{2 x ^{2}} d x

\frac{1}{2} (\frac{x ^{3}}{3} - \frac{1}{x}) + C

Solution steps

\int \frac{x ^{4} + 1}{2 x ^{2}} d x

Take the constant out:

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

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

Expand

\frac{x ^{4} + 1}{x ^{2}} : x^{2} + \frac{1}{x ^{2}}

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

\int x^{2} d x = \frac{x ^{3}}{3}

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

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

Add a constant to the solution

= \frac{1}{2} (\frac{x ^{3}}{3} - \frac{1}{x}) + C

\int \frac{x}{x ^{4} + 2} d x

f (x) = \frac{5}{x}, at x = 1

\int t e^{t} d t

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

y = \frac{1}{x ^{6}}, (1, 1)

What is the integral of (x^4+1)/(2x^2) ?
The integral of (x^4+1)/(2x^2) is 1/2 ((x^3)/3-1/x)+C