What is the integral of (y^2+28y+196)/((y^2+196)^2) ?

The integral of (y^2+28y+196)/((y^2+196)^2) is 1/14 arctan(y/(14))-(14)/(y^2+196)+C

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integral of (y^2+28y+196)/((y^2+196)^2)

\int \frac{y ^{2} + 28 y + 196}{( y ^{2} + 196 ) ^{2}} d y

\frac{1}{14} arctan (\frac{y}{14}) - \frac{14}{y ^{2} + 196} + C

Solution steps

\int \frac{y ^{2} + 28 y + 196}{( y ^{2} + 196 ) ^{2}} d y

Take the partial fraction of

\frac{y ^{2} + 28 y + 196}{( y ^{2} + 196 ) ^{2}} : \frac{1}{y ^{2} + 196} + \frac{28 y}{( y ^{2} + 196 ) ^{2}}

= \int \frac{1}{y ^{2} + 196} + \frac{28 y}{( y ^{2} + 196 ) ^{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} + 196} d y + \int \frac{28 y}{( y ^{2} + 196 ) ^{2}} d y

\int \frac{1}{y ^{2} + 196} d y = \frac{1}{14} arctan (\frac{y}{14})

\int \frac{28 y}{( y ^{2} + 196 ) ^{2}} d y = - \frac{14}{y ^{2} + 196}

= \frac{1}{14} arctan (\frac{y}{14}) - \frac{14}{y ^{2} + 196}

Add a constant to the solution

= \frac{1}{14} arctan (\frac{y}{14}) - \frac{14}{y ^{2} + 196} + C

x \to - 4 lim (\frac{x ^{2} - 1}{4 - x})

\int_{0}^{1} cos (\frac{π x}{2}) d x

y^{^{''}} + 4 y^{^{'}} = (4 t - 4) sin (3 t)

n = 0 \sum \infty 0. 8^{n}

e^{x} (x + x x)

What is the integral of (y^2+28y+196)/((y^2+196)^2) ?
The integral of (y^2+28y+196)/((y^2+196)^2) is 1/14 arctan(y/(14))-(14)/(y^2+196)+C