Find the integral.
-$\int(1-6 x) e^{\left(3 x-9 x^{2}\right)} d x$
A) $\frac{1}{3} e^{\left(3 x-9 x^{2}\right)}+C$
B) $3(1-6 x) e^{\left(3 x-9 x^{2}\right)}+C$
C) $\frac{1}{3}(1-6 x) e^{\left(3 x-9 x^{2}\right)}+C$
D) $3 e^{\left(3 x-9 x^{2}\right)}+C$

Question

Quizplus · Accepted Answer

The integral can be solved using substitution. Let $ u = 3x - 9x^2 $, then $ du = (3 - 18x) dx $. The expression $ (1 - 6x) $ is a factor of $ (3 - 18x) $, allowing the integral to simplify to $\frac{1}{3} e^u du$, which integrates to $\frac{1}{3} e^{(3x - 9x^2)} + C$.

Find the Integral ∫(1−6x)e(3x−9x2)dx\int(1-6 x) e^{\left(3 x-9 x^{2}\right)} d x∫(1−6x)e(3x−9x2)dx A) 13e(3x−9x2)+C\frac{1}{3} e^{\left(3 x-9 x^{2}\right)}+C31​e(3x−9x2)+C

Find the Integral $\int(1-6 x) e^{\left(3 x-9 x^{2}\right)} d x$ A) $\frac{1}{3} e^{\left(3 x-9 x^{2}\right)}+C$