# Evaluate $\int_{0}^{1/2}\frac{e^x(2-x^2)}{(1-x)^{3/2}(1+x)^{1/2}}\,dx$

Consider the following intergal: $$\int_{0}^{1/2}\frac{e^x(2-x^2)}{(1-x)^{3/2}(1+x)^{1/2}}dx$$

I have tried this by trying to write the numerator as 1 + the product of the below-given functions and then using partial fractions. I've also thought of trying to do some trig substitutions but then that simply gives an exponential raised to a trig ratio etc.

All in all, this integral has beaten me. An online calculator gives the following indefinite value: $$-\frac{\sqrt{1-x}\sqrt{1+x}e^x}{(x-1)} + C$$ While the definite value is: $$\sqrt{3e}-1$$

With the solution, can the answerer also add what came to his mind when he looked at the integral, in the sense of how did he think to approach this problem?

There must obviously be some features of this expression that more experienced people can see and hence can guess the method which might work. I want to know what such features have been identified while solving this. On the other hand, if there is just a normal standard way of solving this. I would love to know that too!

Sometimes, the definite integrals of the form $$\int_a^b\mathrm e^x\cdot F(x)\mathrm dx$$, where $$F(x)$$ is a suitable irrational function, can be solved with the help of a clever per partes application. This is also your case.

We have the following:

\begin{align*} I&:=\int_{0}^{\frac12}\mathrm e^x\cdot\frac{2-x^2}{(1-x)^{\frac32}\cdot(1+x)^{\frac12}}\,\mathrm dx\\[1em] &=\int_{0}^{\frac12}\mathrm e^x\cdot\frac{1}{(1-x)^{\frac32}\cdot(1+x)^{\frac12}}\mathrm dx+{\color{red}{\int_{0}^{\frac12}\mathrm e^x\cdot\frac{1-x^2}{(1-x)^{\frac32}\cdot(1+x)^{\frac12}}\mathrm dx}}. \end{align*}

Since the derivative of $$(1-x^2)\Bigl/\left[(1-x)^{\frac32}\cdot(1-x)^{\frac12}\right]$$ equals to $$1\Bigl/\left[(1-x)^{\frac32}\cdot(1-x)^{\frac12}\right]$$, one can discover the connection to the first integral above. This gives us some hope of success. In virtue of this, we obtain

\begin{align*} I&=\int_{0}^{\frac12}\mathrm e^x\cdot\frac{1}{(1-x)^{\frac32}\cdot(1+x)^{\frac12}}\mathrm dx+{\color{red}{\mathrm e^x\cdot\left.\frac{1-x^2}{(1-x)^{\frac32}\cdot(1+x)^{\frac12}}\right |_{0}^{\frac12}-\int_{0}^{\frac12}\mathrm e^x\cdot\frac{1}{(1-x)^{\frac32}\cdot(1+x)^{\frac12}}\mathrm dx}}\\[1em] &=e^x\cdot\left.\frac{1-x^2}{(1-x)^{\frac32}\cdot(1+x)^{\frac12}}\right |_{0}^{\frac12}=e^x\cdot\left.\sqrt{\frac{1+x}{1-x}}\right |_{0}^{\frac12}\\[1em] &=\mathrm e^{\frac12}\cdot\sqrt{\frac{\frac32}{\frac12}}-1=\boldsymbol{\sqrt{3\mathrm e}-1}, \end{align*}

as you claim.

Recall the general exponential integral result

$$\int [f(x)+f’(x)]e^x dx= f(x) e^x$$ and note that

\begin{align} &\frac{2-x^2}{(1-x)^{3/2}(1+x)^{1/2}} =\sqrt{\frac{1+x}{1-x} }+ \frac{1}{(1-x)^{3/2}(1+x)^{1/2}}= f(x) + f’(x)\\ \end{align}

where $$f(x)= \sqrt{\frac{1+x}{1-x} }$$. Thus

$$\int_{0}^{1/2}\frac{e^x(2-x^2)}{(1-x)^{3/2}(1+x)^{1/2}}dx =\sqrt{\frac{1+x}{1-x} }\>e^x\bigg|_0^{1/2}=\sqrt{3e}-1$$

Here's a slightly different take on this problem using a differential equation approach.

As the integrand contains a factor of $$e^x$$ we expect the result to contain the same. Write $$\frac{e^x(2-x^2)}{(1-x)^{3/2}(1+x)^{1/2}}=(e^xf(x))'\implies\frac{2-x^2}{(1-x)\sqrt{1-x^2}}=f'(x)+f(x).$$ To put the LHS into polynomial form, make the substitution $$f(x)=g(x)\sqrt{1-x^2}$$. This gives $$g'(x)\sqrt{1-x^2}-\frac{xg(x)}{\sqrt{1-x^2}}+g(x)\sqrt{1-x^2}=\frac{2-x^2}{(1-x)\sqrt{1-x^2}}$$ so that $$(1-x^2)(1-x)g'(x)+(1-x-x^2)(1-x)g(x)=2-x^2$$.

Consider the power series $$g(x)=a_0+a_1x+a_2x^2+\cdots$$ so \begin{align}2-x^2&=(1-x-x^2+x^3)(a_1+2a_2x+3a_3x^2+\cdots)\\&\quad+(1-2x+x^3)(a_0+a_1x+a_2x^2+\cdots).\end{align} Equating constant terms yields $$a_1=2-a_0$$ and equating linear terms yields $$a_2=a_0$$. Equating quadratic terms yields $$3a_3=a_2+3a_1-1=5-2a_0$$ and equating cubic terms yields $$3a_4=1+2a_0$$. Notice that taking $$a_0=1$$ yields $$a_1=a_2=a_3=1$$.

In general, equating the coefficients of $$x^k$$ for $$k>3$$ on both sides yields $$0=(k+1)a_{k+1}-(k-1)a_k-(k+1)a_{k-1}+(k-2)a_{k-2}+a_{k-3}.$$ Substituting $$k=4$$ yields $$a_4=1$$, and by simple induction, it follows that $$a_i=1$$ for all $$i\ge0$$. Hence $$g(x)=1/(1-x)$$ from which $$\int\frac{e^x(2-x^2)}{(1-x)^{3/2}(1+x)^{1/2}}\,dx=e^x\cdot\frac1{1-x}\cdot\sqrt{1-x^2}+C=e^x\sqrt{\frac{1+x}{1-x}}+C.$$