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

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

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

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

arctan (x) + x + C

Solution steps

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

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

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

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

\int 1 d x = x

= arctan (x) + x

Add a constant to the solution

= arctan (x) + x + C

\frac{\partial}{\partial x} (ln (x + 5 y + 6 z))

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x^{3} - 3 x + 4

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

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