Let f(x) = (2-x^2)e^x√1+x,(1-x)^3/2,dx, with f(0)=0.

Question

Let $f(x) = \displaystyle\int \frac{(2-x^2)e^x}{\sqrt{1+x}\,(1-x)^{3/2}}\,dx$, with $f(0)=0$. Then $f\left(\dfrac12\right)$ is equal to:

Accepted Answer

$$f(x) = \int\frac{(2-x^2)e^x}{\sqrt{1-x^2}\,(1-x)}\,dx = \int e^x\left[\frac{1}{\sqrt{1-x^2}(1-x)} + \frac{1+x}{\sqrt{1-x^2}}\right]dx$$

Let $g(x) = \dfrac{1+x}{\sqrt{1-x^2}}$; one verifies $g'(x) = \dfrac{1}{\sqrt{1-x^2}(1-x)}$, so the integrand is $e^x[g(x)+g'(x)]$:

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

Using $f(0)=0$: $1+C=0 \implies C=-1$. So $f(x) = \dfrac{e^x(1+x)}{\sqrt{1-x^2}} - 1$.

$$f\left(\frac12\right) = \frac{\sqrt e\cdot\frac32}{\sqrt{3}/2} - 1 = \sqrt e\cdot\frac{3}{\sqrt3}-1 = \sqrt{3e}-1$$

Solutions · 1