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

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

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

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

arctan (x) + \frac{1}{2} ln 1 + x^{2} + C

Solution steps

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

Expand

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

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

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

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

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

Add a constant to the solution

= arctan (x) + \frac{1}{2} ln 1 + x^{2} + C

x = - 2, x = 1, y = 2 x^{2} + 12, y = 0

f (x) = 5 x + 11, (5, 6)

f (x) = \frac{3}{x ^{2}}

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

\int (ln (x^{3})) d x

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