What is the integral of (y^2+34y+289)/((y^2+289)^2) ?

The integral of (y^2+34y+289)/((y^2+289)^2) is 1/17 arctan(y/(17))-(17)/(y^2+289)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (y^2+34y+289)/((y^2+289)^2)

\int \frac{y ^{2} + 34 y + 289}{( y ^{2} + 289 ) ^{2}} d y

\frac{1}{17} arctan (\frac{y}{17}) - \frac{17}{y ^{2} + 289} + C

Solution steps

\int \frac{y ^{2} + 34 y + 289}{( y ^{2} + 289 ) ^{2}} d y

Take the partial fraction of

\frac{y ^{2} + 34 y + 289}{( y ^{2} + 289 ) ^{2}} : \frac{1}{y ^{2} + 289} + \frac{34 y}{( y ^{2} + 289 ) ^{2}}

= \int \frac{1}{y ^{2} + 289} + \frac{34 y}{( y ^{2} + 289 ) ^{2}} d y

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{1}{y ^{2} + 289} d y + \int \frac{34 y}{( y ^{2} + 289 ) ^{2}} d y

\int \frac{1}{y ^{2} + 289} d y = \frac{1}{17} arctan (\frac{y}{17})

\int \frac{34 y}{( y ^{2} + 289 ) ^{2}} d y = - \frac{17}{y ^{2} + 289}

= \frac{1}{17} arctan (\frac{y}{17}) - \frac{17}{y ^{2} + 289}

Add a constant to the solution

= \frac{1}{17} arctan (\frac{y}{17}) - \frac{17}{y ^{2} + 289} + C

(csc (x) + cot (x))^{- 1}

x \to \infty lim (6)

\frac{d}{d t} (\frac{t + 1}{t - 1})

x, x^{\frac{1}{2}}, 1, 2

f (x) = 3 sin (x), at x = \frac{π}{3}

What is the integral of (y^2+34y+289)/((y^2+289)^2) ?
The integral of (y^2+34y+289)/((y^2+289)^2) is 1/17 arctan(y/(17))-(17)/(y^2+289)+C