What is the integral of (y^3+4y)/(9+y^4) ?

The integral of (y^3+4y)/(9+y^4) is 1/4 ln|9+y^4|+2/3 arctan((y^2)/3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (y^3+4y)/(9+y^4)

\int \frac{y ^{3} + 4 y}{9 + y ^{4}} d y

\frac{1}{4} ln 9 + y^{4} + \frac{2}{3} arctan (\frac{y ^{2}}{3}) + C

Solution steps

\int \frac{y ^{3} + 4 y}{9 + y ^{4}} d y

Expand

\frac{y ^{3} + 4 y}{9 + y ^{4}} : \frac{y ^{3}}{9 + y ^{4}} + \frac{4 y}{9 + y ^{4}}

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{y ^{3}}{9 + y ^{4}} d y + \int \frac{4 y}{9 + y ^{4}} d y

\int \frac{y ^{3}}{9 + y ^{4}} d y = \frac{1}{4} ln 9 + y^{4}

\int \frac{4 y}{9 + y ^{4}} d y = \frac{2}{3} arctan (\frac{y ^{2}}{3})

= \frac{1}{4} ln 9 + y^{4} + \frac{2}{3} arctan (\frac{y ^{2}}{3})

Add a constant to the solution

= \frac{1}{4} ln 9 + y^{4} + \frac{2}{3} arctan (\frac{y ^{2}}{3}) + C

\int (\frac{ln ( 53 )}{v} - ln (6) v^{e}) d v

\int \frac{arcsin ( x )}{3 x ^{2}} d x

\int \frac{1}{t ^{2} + 2 t} d t

\int \frac{tan ( x ) + sec ( x )}{sec ^{2} ( x )} d x

\int sin (\frac{θ}{2}) tan (\frac{θ}{2}) d θ

What is the integral of (y^3+4y)/(9+y^4) ?
The integral of (y^3+4y)/(9+y^4) is 1/4 ln|9+y^4|+2/3 arctan((y^2)/3)+C