What is the integral of (3+x^2)/(1+x^2) ?

The integral of (3+x^2)/(1+x^2) is 2arctan(x)+x+C

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integral of (3+x^2)/(1+x^2)

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

2 arctan (x) + x + C

Solution steps

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

\frac{3 + x ^{2}}{1 + x ^{2}} = \frac{2}{1 + x ^{2}} + 1

= \int \frac{2}{1 + x ^{2}} + 1 d x

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{2}{1 + x ^{2}} d x + \int 1 d x

\int \frac{2}{1 + x ^{2}} d x = 2 arctan (x)

\int 1 d x = x

= 2 arctan (x) + x

Add a constant to the solution

= 2 arctan (x) + x + C

\frac{\partial}{\partial y} (x + y)

x, 100

\frac{d y}{d x} = x + 2 y, y (0) = 7

f (x) = 2 - 7 x^{2}, at x = 4

\frac{d}{d x} (e^{x} \cdot cos (y))

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