Find the indefinite integral.
$$\int \left( x ^ { 2 } + 4 x - 5 \right) ^ { 2 } ( 2 x + 4 ) d x$$
A) $- 2 \left( x ^ { 2 } + 4 x - 5 \right) ^ { 3 } + C$
B) $2 \left( x ^ { 2 } + 4 x - 5 \right) ^ { 3 } + C$
C) $\left( x ^ { 2 } + 4 x - 5 \right) ^ { 3 } + C$
D) $\frac { 1 } { 3 } \left( x ^ { 2 } + 4 x - 5 \right) ^ { 3 } + C$

Question

Quizplus · Accepted Answer

We can first expand the expression inside the integral using the binomial theorem:
\begin{align*}
\left( x ^ { 2 } + 4 x - 5 ight) ^ { 2 } ( 2 x + 4 ) &= \left( (x^2)^2 + 2(x^2)(4x) + (4x)^2 ight) (2x + 4) - 5^2(2x+4) \
&= (2x+4)\left( 2x^4 + 16x^3 + 36x^2 - 50x - 75 ight)
\end{align*}
Now we can integrate term by term:
\begin{align*}
\int \left( x ^ { 2 } + 4 x - 5 ight) ^ { 2 } ( 2 x + 4 ) d x &=  \int (2x+4)\left( 2x^4 + 16x^3 + 36x^2 - 50x - 75 ight) dx \
&= \int 4x^5 + 32x^4 + 72x^3 - 100x^2 - 150x dx + \int 8x^4 + 64x^3 + 144x^2 - 200x - 300 dx \
&= \frac { 4 } { 6 } x ^ { 6 } + \frac { 32 } { 5 } x ^ { 5 } + 18x^4 - \frac { 100 } { 3 } x ^ { 3 } - \frac { 75 } { 2 } x ^ { 2 } + \frac { 8 } { 5 } x ^ { 5 } + 16x^4 + \frac { 144 } { 3 } x ^ { 3 } - 100x^2 - 150x + C \
&= \frac { 2 } { 3 } x ^ { 6 } + \frac { 252 } { 15 } x ^ { 5 } + 34x^4 - \frac { 200 } { 3 } x ^ { 3 } - \frac { 475 } { 6 } x ^ { 2 } - 170x + C\
&= \frac { 1 } { 3 } \left( x ^ { 2 } + 4 x - 5 ight) ^ { 3 } + C
\end{align*}
Therefore, the correct choice is D.

Find the Indefinite Integral ∫(x2+4x−5)2(2x+4)dx\int \left( x ^ { 2 } + 4 x - 5 \right) ^ { 2 } ( 2 x + 4 ) d x∫(x2+4x−5)2(2x+4)dx

Find the Indefinite Integral $\int \left( x ^ { 2 } + 4 x - 5 \right) ^ { 2 } ( 2 x + 4 ) d x$