Answer
$$\int\frac{1+x}{1+x^2}dx=\arctan x+\frac{1}{2}\ln(1+x^2)+c$$
Work Step by Step
We will first separate this integral into two new ones by additivity of integrals:
$$\int\frac{1+x}{1+x^2}dx=\int\frac{1}{1+x^2}dx+\int\frac{x}{1+x^2}dx$$
First integral:
$$\int\frac{1}{1+x^2}dx=\arctan x+c$$
To evaluate the second integral we will use substitution $1+x^2=t$ which gives us $2xdx=dt\Rightarrow xdx=\frac{dt}{2}$. Putting this into the integral we get:
$$\int\frac{x}{1+x^2}dx=\int\frac{1}{t}\frac{dt}{2}=\frac{1}{2}\ln|t|+c$$
Now we have to express solution in terms of $x$:
$$\int\frac{x}{1+x^2}dx=\frac{1}{2}\ln|t|+c=\frac{1}{2}\ln(1+x^2)+c$$
Finally we have:
$$\int{1+x}{1+x^2}dx=\arctan x+\frac{1}{2}\ln(1+x^2)+c$$
