What is the integral of ((y^2+1))/(y^2+4) ?

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

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integral of ((y^2+1))/(y^2+4)

\int \frac{( y ^{2} + 1 )}{y ^{2} + 4} d y

- \frac{3}{2} arctan (\frac{y}{2}) + y + C

Solution steps

\int \frac{y ^{2} + 1}{y ^{2} + 4} d y

\frac{y ^{2} + 1}{y ^{2} + 4} = - \frac{3}{y ^{2} + 4} + 1

= \int - \frac{3}{y ^{2} + 4} + 1 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{3}{y ^{2} + 4} d y + \int 1 d y

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

\int 1 d y = y

= - \frac{3}{2} arctan (\frac{y}{2}) + y

Add a constant to the solution

= - \frac{3}{2} arctan (\frac{y}{2}) + y + C

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

\frac{5}{4 ( s - 3 ) ^{2} - 64}

\frac{d}{d x} (ln (\frac{x + 2}{x ^{2}}))

\frac{d}{d x} (x^{2} y^{4})

x \to 2 lim (x^{2} - 5)

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