What is the integral of (y^2+26y+169)/((y^2+169)^2) ?

The integral of (y^2+26y+169)/((y^2+169)^2) is 1/13 arctan(y/(13))-(13)/(y^2+169)+C

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integral of (y^2+26y+169)/((y^2+169)^2)

\int \frac{y ^{2} + 26 y + 169}{( y ^{2} + 169 ) ^{2}} d y

\frac{1}{13} arctan (\frac{y}{13}) - \frac{13}{y ^{2} + 169} + C

Solution steps

\int \frac{y ^{2} + 26 y + 169}{( y ^{2} + 169 ) ^{2}} d y

Take the partial fraction of

\frac{y ^{2} + 26 y + 169}{( y ^{2} + 169 ) ^{2}} : \frac{1}{y ^{2} + 169} + \frac{26 y}{( y ^{2} + 169 ) ^{2}}

= \int \frac{1}{y ^{2} + 169} + \frac{26 y}{( y ^{2} + 169 ) ^{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} + 169} d y + \int \frac{26 y}{( y ^{2} + 169 ) ^{2}} d y

\int \frac{1}{y ^{2} + 169} d y = \frac{1}{13} arctan (\frac{y}{13})

\int \frac{26 y}{( y ^{2} + 169 ) ^{2}} d y = - \frac{13}{y ^{2} + 169}

= \frac{1}{13} arctan (\frac{y}{13}) - \frac{13}{y ^{2} + 169}

Add a constant to the solution

= \frac{1}{13} arctan (\frac{y}{13}) - \frac{13}{y ^{2} + 169} + C

\int (cos (2 x) + sin (2 x))^{2} d x

(ln (y))^{7} \frac{d y}{d x} = x^{7} y, y (1) = e^{2}

\frac{( x ^{3} - 5 x + 186 )}{x ^{2} - x - 42}

sin (x), cos (x), \frac{- π}{30}, \frac{25}{12}

y^{^{''}} + y^{^{'}} + 1.25 y = 0, y (0) = 3, y^{^{'}} (0) = 1

What is the integral of (y^2+26y+169)/((y^2+169)^2) ?
The integral of (y^2+26y+169)/((y^2+169)^2) is 1/13 arctan(y/(13))-(13)/(y^2+169)+C