Example 3.4. Find the indefinite integral

$$
\int \frac{(x+1) d x}{\sqrt{3-x^{2}}}
$$

and verify the result by differentiation.
Solution. We decompose the given integral into the sum of two integrals:

$$
\int \frac{(x+1) d x}{\sqrt{3-x^{2}}}=\int \frac{x d x}{\sqrt{3-x^{2}}}+\int \frac{d x}{\sqrt{3-x^{2}}}
$$

To compute the first of these integrals, we use the fact that

$$
x d x=\frac{1}{2} d\left(x^{2}\right)=-\frac{1}{2} d\left(3-x^{2}\right)
$$

(thus, the factor $x$ is "brought under the differential sign"), and we make the substitution:

$$
t=3-x^{2} .
$$

We obtain:

$$
\int \frac{x d x}{\sqrt{3-x^{2}}}=-\frac{1}{2} \int \frac{d\left(3-x^{2}\right)}{\sqrt{3-x^{2}}}=-\frac{1}{2} \int^{-\frac{1}{2}} d t .
$$

The obtained integral is a standard one: $\int^{\alpha} d t=\frac{t^{\alpha+1}}{\alpha+1}+C$.

Applying this formula for $\alpha=-\frac{1}{2}$, we have: $-\frac{1}{2} \int t^{-\frac{1}{2}} d t=-\sqrt{t}+C$.

Returning to the variable $x$, we get:

$$
\int \frac{x d x}{\sqrt{3-x^{2}}}=-\sqrt{3-x^{2}}+C
$$

Similarly, we compute the integral

$$
\int \frac{d x}{\sqrt{3-x^{2}}}=\frac{1}{\sqrt{3}} \int d \arcsin \frac{x}{\sqrt{3}}=\arcsin \frac{x}{\sqrt{3}}+C
$$

Finally, we have:

$$
\int \frac{(x+1) d x}{\sqrt{3-x^{2}}}=-\sqrt{3-x^{2}}+\arcsin \frac{x}{\sqrt{3}}+C
$$

Verification. Let's ensure that the derivative of the obtained expression matches the integrand. Applying the table of derivatives and the rule for differentiating composite functions, we find:

$$
\begin{gathered}
(-\sqrt{3-x^{2}})=-\frac{1}{2} \frac{1}{\sqrt{3-x^{2}}}\left(3-x^{2}\right)^{\prime}=\frac{x}{\sqrt{3-x^{2}}} \\
\arcsin \frac{x}{\sqrt{3}}=\frac{1}{\sqrt{1-\left(\frac{x}{\sqrt{3}}\right)^{2}}} \frac{1}{\sqrt{3}}=\frac{1}{\sqrt{3-x^{2}}}
\end{gathered}
$$

Adding these two expressions, we obtain the integrand. Therefore, the integration is performed correctly.