We are given the integral \( \int \frac{4x^9}{x^{10} - 10} \, dx \) and are asked to find the solution.
We can solve this by using substitution. Let:
\( u = x^{10} - 10 \),
so that \( du = 10x^9 \, dx \), or equivalently \( \frac{du}{10} = x^9 \, dx \).
Now, rewrite the integral in terms of \( u \):
\( \int \frac{4x^9}{x^{10} - 10} \, dx = \frac{4}{10} \int \frac{1}{u} \, du = \frac{2}{5} \int \frac{1}{u} \, du \).
The integral of \( \frac{1}{u} \) is \( \log|u| \), so we have:
\( \frac{2}{5} \log|u| + C \).
Substitute back \( u = x^{10} - 10 \):
\( \frac{2}{5} \log|x^{10} - 10| + C \).
The correct answer is \( \frac{2}{5} \log|x^{10} - 10| + C \).
Draw a rough sketch for the curve $y = 2 + |x + 1|$. Using integration, find the area of the region bounded by the curve $y = 2 + |x + 1|$, $x = -4$, $x = 3$, and $y = 0$.
If the function \[ f(x) = \begin{cases} \frac{2}{x} \left( \sin(k_1 + 1)x + \sin(k_2 -1)x \right), & x<0 \\ 4, & x = 0 \\ \frac{2}{x} \log_e \left( \frac{2 + k_1 x}{2 + k_2 x} \right), & x>0 \end{cases} \] is continuous at \( x = 0 \), then \( k_1^2 + k_2^2 \) is equal to: