Final answer to the problem
$\int{\left(\left(-\frac{33}{10}\right)\right)}^{-141}\cdot {\left(\left(-\frac{33}{10}\right)\right)}^3\cdot {\left(\left(-\frac{33}{10}\right)\right)}^{-121}d$
Got another answer? Verify it here!
Step-by-step Solution
1
Find the integral
$\int\left(11\cdot - \left(\frac{3}{10}\right)\right)^{141\cdot -1}\cdot \left(11\cdot - \left(\frac{3}{10}\right)\right)^3\cdot \left(11\cdot - \left(\frac{3}{10}\right)\right)^{121\cdot -1}d$
2
Simplifying
$\int{\left(\left(-\frac{33}{10}\right)\right)}^{-141}\cdot {\left(\left(-\frac{33}{10}\right)\right)}^3\cdot {\left(\left(-\frac{33}{10}\right)\right)}^{-121}d$
Final answer to the problem
$\int{\left(\left(-\frac{33}{10}\right)\right)}^{-141}\cdot {\left(\left(-\frac{33}{10}\right)\right)}^3\cdot {\left(\left(-\frac{33}{10}\right)\right)}^{-121}d$