Question

Evaluate:$\int \left[(x^2+1)\sqrt{(x+1)}\right]dx$

• A. $\begin{array}{l}\frac{(x+1{)}^{7/2}}{7}-2\frac{(x+1{)}^{5/2}}{5}+2\frac{(x+1{)}^{3/2}}{3}+C\end{array}$
• B. $\begin{array}{l}2\left[\frac{(x+1{)}^{7/2}}{7}-2\frac{(x+1{)}^{5/2}}{5}+2\frac{(x+1{)}^{3/2}}{3}\right]+C\end{array}$
• C. $\begin{array}{l}\frac{(x+1{)}^{7/2}}{7}-2\frac{(x+1{)}^{5/2}}{5}+5\end{array}$
• D. $\begin{array}{l}\frac{(x+1{)}^{7/2}}{7}-3\frac{(x+1{)}^{5/2}}{5}+1(x+1{)}^{1/2}+C\end{array}$
• E. $\begin{array}{l}(x+1{)}^{7/2}+(x+1{)}^{5/2}+(x+1{)}^{3/2}+C\end{array}$

Answer 1

The correct option is B

$\begin{array}{l}2\left[\frac{(x+1{)}^{7/2}}{7}-2\frac{(x+1{)}^{5/2}}{5}+2\frac{(x+1{)}^{3/2}}{3}\right]+C\end{array}$

Evaluating the integral$\int \left[(x^2+1)\sqrt{(x+1)}\right]dx$

Substituting $t^2=(x+1)$ and $dx=2dt$we get

$$\begin{gathered} \Rightarrow \int (t^4-2t^2+2)t2tdt\left[\because x^2+1=t^4-2t^2+2\right] \\ \Rightarrow 2\int (t^6-2t^4+2t^2)dt \\ \Rightarrow 2\left[\int t^{6}dt-2\int t^4+2\int t^{2}dt\right] \\ \Rightarrow 2\left[\frac{t^7}{7}-2\frac{t^5}{5}+2\frac{t^3}{3}\right]+C \\ \Rightarrow 2\left[\frac{(x+1{)}^{\frac{7}{2}}}{7}-\frac{2(x+1{)}^{\frac{5}{2}}}{5}+\frac{2(x+1{)}^{\frac{3}{2}}}{3}\right]+C[substitutingback{t}^2=(x+1\left)\right] \end{gathered}$$

Hence, option B is the correct answer.