Question

$\int{(x+1){{(x+2)}^{7}}}(x+3)dx$ is equal to

• A. $\frac{{{(x+2)}^{10}}}{10}-\frac{{{(x+2)}^{8}}}{8}+C$
• B. $\frac{{{(x+1)}^{2}}}{2}-\frac{{{(x+2)}^{8}}}{8}-\frac{{{(x+3)}^{2}}}{2}+C$
• C. $\frac{{{(x+2)}^{10}}}{10}+C$
• D. $\frac{{{(x+1)}^{2}}}{2}+\frac{{{(x+2)}^{8}}}{8}+\frac{{{(x+3)}^{2}}}{2}+C$

Answer 1

Answer: (A)

$\frac{{{(x+2)}^{10}}}{10}-\frac{{{(x+2)}^{8}}}{8}+C$

Answer 2

Let $I=\int{(x+1){{(x+2)}^{7}}(x+3)}dx$
Putting $x+2=t$
$\Rightarrow$ $dx=dt$
Also, $x+1=t-1$ and $x+3=t+1$
$\therefore$ $I=\int{(t-1){{t}^{7}}}(t+1)dt$
$=\int{({{t}^{2}}-1)}{{t}^{7}}dt$
$=\int{{{t}^{9}}}dt-\int{{{t}^{7}}}dt$
$=\frac{{{t}^{10}}}{10}-\frac{{{t}^{8}}}{8}+c$
$=\frac{{{(x+2)}^{10}}}{10}-\frac{{{(x+2)}^{8}}}{8}+c$