What is the integral of (y^2+14y+49)/((y^2+49)^2) ?

The integral of (y^2+14y+49)/((y^2+49)^2) is 1/7 arctan(y/7)-7/(y^2+49)+C

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integral of (y^2+14y+49)/((y^2+49)^2)

\int \frac{y ^{2} + 14 y + 49}{( y ^{2} + 49 ) ^{2}} d y

\frac{1}{7} arctan (\frac{y}{7}) - \frac{7}{y ^{2} + 49} + C

Solution steps

\int \frac{y ^{2} + 14 y + 49}{( y ^{2} + 49 ) ^{2}} d y

Take the partial fraction of

\frac{y ^{2} + 14 y + 49}{( y ^{2} + 49 ) ^{2}} : \frac{1}{y ^{2} + 49} + \frac{14 y}{( y ^{2} + 49 ) ^{2}}

= \int \frac{1}{y ^{2} + 49} + \frac{14 y}{( y ^{2} + 49 ) ^{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} + 49} d y + \int \frac{14 y}{( y ^{2} + 49 ) ^{2}} d y

\int \frac{1}{y ^{2} + 49} d y = \frac{1}{7} arctan (\frac{y}{7})

\int \frac{14 y}{( y ^{2} + 49 ) ^{2}} d y = - \frac{7}{y ^{2} + 49}

= \frac{1}{7} arctan (\frac{y}{7}) - \frac{7}{y ^{2} + 49}

Add a constant to the solution

= \frac{1}{7} arctan (\frac{y}{7}) - \frac{7}{y ^{2} + 49} + C

y^{^{'}} + (1 + 3 x^{2}) y = 3 + 9 x^{2}

\int \frac{1 + 2 t ^{3}}{t} d t

\frac{\partial}{\partial x} (2 e^{2 x})

f (x) = x (x^{3} + 2 x + 3)

y^{^{''}} = a^{2} y

What is the integral of (y^2+14y+49)/((y^2+49)^2) ?
The integral of (y^2+14y+49)/((y^2+49)^2) is 1/7 arctan(y/7)-7/(y^2+49)+C