What is the integral of (x^4+x^2+25)/(5x^4(x^2+25)) ?

The integral of (x^4+x^2+25)/(5x^4(x^2+25)) is 1/5 (-1/(3x^3)+1/5 arctan(x/5))+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+x^2+25)/(5x^4(x^2+25))

\int \frac{x ^{4} + x ^{2} + 25}{5 x ^{4} ( x ^{2} + 25 )} d x

\frac{1}{5} (- \frac{1}{3 x ^{3}} + \frac{1}{5} arctan (\frac{x}{5})) + C

Solution steps

\int \frac{x ^{4} + x ^{2} + 25}{5 x ^{4} ( x ^{2} + 25 )} d x

Take the constant out:

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

= \frac{1}{5} \cdot \int \frac{x ^{4} + x ^{2} + 25}{x ^{4} ( x ^{2} + 25 )} d x

Take the partial fraction of

\frac{x ^{4} + x ^{2} + 25}{x ^{4} ( x ^{2} + 25 )} : \frac{1}{x ^{4}} + \frac{1}{x ^{2} + 25}

= \frac{1}{5} \cdot \int \frac{1}{x ^{4}} + \frac{1}{x ^{2} + 25} 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}{5} (\int \frac{1}{x ^{4}} d x + \int \frac{1}{x ^{2} + 25} d x)

\int \frac{1}{x ^{4}} d x = - \frac{1}{3 x ^{3}}

\int \frac{1}{x ^{2} + 25} d x = \frac{1}{5} arctan (\frac{x}{5})

= \frac{1}{5} (- \frac{1}{3 x ^{3}} + \frac{1}{5} arctan (\frac{x}{5}))

Add a constant to the solution

= \frac{1}{5} (- \frac{1}{3 x ^{3}} + \frac{1}{5} arctan (\frac{x}{5})) + C

\frac{\partial}{\partial x} (2 x y (x + y) + x^{2} + 2 x)

\frac{\partial}{\partial y} (ln (x^{2} y))

7 \cdot ln (x)

sin^{3} (3 x)

\frac{\partial}{\partial y} (- 2 cos (y))

What is the integral of (x^4+x^2+25)/(5x^4(x^2+25)) ?
The integral of (x^4+x^2+25)/(5x^4(x^2+25)) is 1/5 (-1/(3x^3)+1/5 arctan(x/5))+C