## Problem 3

Calculate: $\int \frac{x+1-x^{2} \cdot \ln x}{x^{3}+x^{2}} \cdot \cos [\ln (x+1)] d x, x>0$.
## Solution:

$$
\int \frac{x+1-x^{2} \cdot \ln x}{x^{2}(x+1)} \cdot \cos [\ln (x+1)] d x=\int \frac{1}{x^{2}} \cdot \cos [\ln (x+1)] d x-\int \frac{\ln x}{x+1} \cdot \cos [\ln (x+1)] d x=
$$

For any other correct solution, full credit is given

$$
\begin{aligned}
& =\int\left(-\frac{1}{x}\right)^{\prime} \cdot \cos [\ln (x+1)] d x-\int \ln x \cdot[\sin \ln (x+1)]^{\prime} d x= \\
& =-\frac{1}{x} \cdot \cos [\ln (x+1)]-\int \frac{1}{x} \cdot \sin \ln (x+1) \cdot \frac{1}{x+1} d x- \\
& -\ln x \cdot \sin \ln (x+1)+\int \frac{1}{x} \cdot \sin \ln (x+1) d x= \\
& =-\frac{1}{x} \cdot \cos [\ln (x+1)]-\int\left(\frac{1}{x}-\frac{1}{x+1}\right) \cdot \sin \ln (x+1) d x- \\
& -\ln x \cdot \sin \ln (x+1)+\int \frac{1}{x} \cdot \sin \ln (x+1) d x= \\
& =-\frac{1}{x} \cdot \cos [\ln (x+1)]-\int \frac{1}{x} \cdot \sin \ln (x+1) d x+\int \frac{1}{x+1} \cdot \sin \ln (x+1) d x- \\
& -\ln x \cdot \sin \ln (x+1)+\int \frac{1}{x} \cdot \sin \ln (x+1) d x= \\
& =-\frac{1}{x} \cdot \cos [\ln (x+1)]-\cos [\ln (x+1)]-\ln x \cdot \sin [\ln (x+1)]+C
\end{aligned}
$$

By default: $1 \mathbf{p}$