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

The integral of (6x^5+2x)/(x^6+x^2) is ln|x^4+1|+2ln(x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

ln x^{4} + 1 + 2 ln (x) + C

Solution steps

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

Expand

\frac{6 x ^{5} + 2 x}{x ^{6} + x ^{2}} : \frac{6 x ^{5}}{x ^{6} + x ^{2}} + \frac{2 x}{x ^{6} + x ^{2}}

Apply the Sum Rule:

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

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

\int \frac{6 x ^{5}}{x ^{6} + x ^{2}} d x = \frac{3}{2} ln x^{4} + 1

\int \frac{2 x}{x ^{6} + x ^{2}} d x = \frac{1}{2} (4 ln (x) - ln x^{4} + 1)

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

Simplify

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

= ln x^{4} + 1 + 2 ln (x)

Add a constant to the solution

= ln x^{4} + 1 + 2 ln (x) + C

\frac{d}{d x} (x^{\frac{4}{9}})

\int_{6}^{8} \frac{32}{( x - 6 ) ^{3}} d x

\frac{2 x}{x - 3}

y^{^{'}} + y = e^{x} y^{2}

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

What is the integral of (6x^5+2x)/(x^6+x^2) ?
The integral of (6x^5+2x)/(x^6+x^2) is ln|x^4+1|+2ln(x)+C