Let f(x)= (2-x^2) e^x(√1+x)(1-x)^3/2dx.

Question

Let $f(x)=\int \frac{(2-{x}^{2})\cdot {e}^{x}}{(\sqrt{1+x})(1-x{)}^{3/2}}dx$. If $f(0)=0$, then $f(\frac{1}{2})$ is equal to:

Accepted Answer

$f(x)=\int \frac{(2-{x}^{2})\cdot {e}^{x}}{(\sqrt{1+x})(1-x{)}^{3/2}}dx$

$\Rightarrow$ $f(x)=\int \frac{(2-{x}^{2}){e}^{x}}{\sqrt{1+x}\sqrt{1-x(1-x)}}dx$

$f(x)=\int \frac{(2-{x}^{2})\cdot {e}^{x}}{(\sqrt{1-{x}^{2}})(1-x)}dx$

$\begin{gathered} =\int \frac{[1+(1-{x}^{2})]{e}^{x}}{(\sqrt{1-{x}^{2}})(1-x)}dx \ =\int {e}^{x}[[|\frac{1}{(\sqrt{1-{x}^{2}})(1-x)}+\frac{1+x}{\sqrt{1-{x}^{2}}}]]|dx \end{gathered}$

This integral of the form $\int {e}^{x}[g(x)+{g}^{'}(x)]dx={e}^{x}\cdot g(x)+C$

Let $g(x)=\frac{1+x}{\sqrt{1-{x}^{2}}}$

${g}^{'}(x)=\frac{(1+x{)}^{'}\cdot \sqrt{1-{x}^{2}}-(1+x){(\sqrt{1-{x}^{2}})}^{'}}{{(\sqrt{1-{x}^{2}})}^{2}}$

$=\frac{\sqrt{1-{x}^{2}}+\frac{({x}^{2}+x)}{\sqrt{1-{x}^{2}}}}{1-{x}^{2}}$

$=\frac{1-{x}^{2}+{x}^{2}+x}{\sqrt{1-{x}^{2}}(1-{x}^{2})}$

$=\frac{(x+1)}{\sqrt{1-{x}^{2}}(1-x)(1+x)}$

$=\frac{1}{(\sqrt{1-{x}^{2}})(1-x)}$

So value of integral

$f(x)=\frac{{e}^{x}\cdot (1+x)}{\sqrt{1-{x}^{2}}}+C$

Apply initial condition $f(0)=0$ to find $C$

$0=\frac{{e}^{0}(1+0)}{\sqrt{1-0}}+C\Rightarrow C=-1$

$\Rightarrow$ $f(x)=\frac{{e}^{x}(1+x)}{\sqrt{1-{x}^{2}}}-1$

$	herefore $ $f(1/2)=\frac{{e}^{1/2}(1+1/2)}{\sqrt{1-(1/2{)}^{2}}}-1$

$=\frac{\sqrt{e}\cdot \frac{3}{2}}{\frac{\sqrt{3}}{2}}-1=\sqrt{3e}-1$

Solutions · 1