Question

integrate: x/1+x^2

Answer 1

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

Answer 2

The problem asks to integrate the function $\frac{x}{1+x^2}$.

To solve this integral, we can use the method of substitution.

Let $u = 1+x^2$.
Now, differentiate $u$ with respect to $x$:
$\frac{du}{dx} = \frac{d}{dx}(1+x^2) = 0 + 2x = 2x$.
From this, we can write $du = 2x \, dx$.

We need to substitute $x \, dx$ in the integral. From $du = 2x \, dx$, we get $x \, dx = \frac{1}{2} du$.

Now, substitute $u$ and $x \, dx$ into the original integral:

$$\int \frac{x}{1+x^2} \, dx = \int \frac{1}{u} \left(\frac{1}{2} du\right)$$

We can pull the constant $\frac{1}{2}$ out of the integral:

$$= \frac{1}{2} \int \frac{1}{u} \, du$$

The integral of $\frac{1}{u}$ with respect to $u$ is $\ln|u|$.

$$= \frac{1}{2} \ln|u| + C$$

Finally, substitute back $u = 1+x^2$:

$$= \frac{1}{2} \ln|1+x^2| + C$$

Since $1+x^2$ is always positive for any real value of $x$ (because $x^2 \ge 0$, so $1+x^2 \ge 1$), we can remove the absolute value sign:

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