Question

Evaluate:$\int (x+1){(x+2)}^7(x+3)dx$

• A. $\left[\frac{{(x+2)}^{10}}{10}\right]–\left[\frac{{(x+2)}^8}{8}\right]+c$
• B. $\left[\frac{{(x+2)}^2}{2}\right]–\left[\frac{{(x+2)}^8}{8}\right]-\left[\frac{{(x+3)}^2}{2}\right]+c$
• C. $\left[\frac{{(x+1)}^{10}}{10}\right]+c$
• D. $\left[\frac{{(x+1)}^2}{2}\right]+\left[\frac{{(x+2)}^8}{8}\right]+\left[\frac{{(x+3)}^2}{2}\right]+c$
• E. $\left[\frac{{(x+2)}^9}{9}\right]–\left[\frac{{(x+2)}^7}{7}\right]+c$

Answer 1

The correct option is A

$\left[\frac{{(x+2)}^{10}}{10}\right]–\left[\frac{{(x+2)}^8}{8}\right]+c$

Evaluating the integral $\int (x+1){(x+2)}^7(x+3)dx$

$\Rightarrow \int (x+2-1){(x+2)}^7(x+2+1)dx$

substituting $t=x+2$ and $dx=dt$ we get

$$\begin{gathered} \Rightarrow \int t^7(t-1)(t+1)dt \\ \Rightarrow \int t^7(t^2-1^2)dt\left[\because (a-b)(a+b)=a^2-b^2\right] \\ \Rightarrow \int \left(t^9-t^7\right)dt \\ \Rightarrow \int t^{9}dt-\int t^{7}dt \\ \Rightarrow \frac{t^{10}}{10}-\frac{t^8}{8}+C[Cisanintegratingcons\tan t] \\ \Rightarrow \frac{{\left(x+2\right)}^{10}}{10}-\frac{{\left(x+2\right)}^8}{8}+C[substitutingbackt=(x+2)] \end{gathered}$$

Hence, option A is the correct answer.