Question

$\frac{1+x}{x \sin^2 (x+\log x)}$

Answer 1

$-\cot (x + \log x) + C$

Answer 2

To evaluate the integral $\int \frac{1+x}{x \sin^2 (x+\log x)} dx$, we can use the method of substitution.

Let $t = x + \log x$.

Now, we find the differential $dt$ by differentiating $t$ with respect to $x$: $dt = \frac{d}{dx}(x + \log x) dx$ $dt = \left(1 + \frac{1}{x}\right) dx$ $dt = \left(\frac{x+1}{x}\right) dx$

Observe that the term $\frac{1+x}{x} dx$ is present in the integrand. So, the integral can be rewritten as: $\int \frac{1}{\sin^2 (x+\log x)} \cdot \left(\frac{1+x}{x}\right) dx$

Substitute $t = x + \log x$ and $dt = \frac{1+x}{x} dx$ into the integral: $\int \frac{1}{\sin^2 t} dt$

We know that $\frac{1}{\sin^2 t} = \csc^2 t$. So, the integral becomes: $\int \csc^2 t dt$

The standard integral of $\csc^2 t$ is $-\cot t + C$, where $C$ is the constant of integration. Thus, $\int \csc^2 t dt = -\cot t + C$

Finally, substitute back $t = x + \log x$: $-\cot (x + \log x) + C$