Question

integration of e^x[(1-x)^2/(1+x)^2]dx​

Answer 1

Answer:

Let T=∫ex(x2+1)(x+1)2dx=∫ex((x+1)2−2x)(x+1)2dx=∫ex((x+1)2−2(x+1)+2)(x+1)2dx=∫ex(x+1)2(x+1)2dx−2∫ex(x+1)(x+1)2dx+2∫ex(x+1)2dx=∫exdx−2∫exx+1dx+2∫ex(x+1)2dx=ex−2(∫exx+1dx−∫ex(x+1)2dx)  

We use integration by parts on the first of these integrals, which uses the following general rule: ∫udv=uv−∫vdu. So let u=1x+1 and v=ex.

∴∫exx+1dx=exx+1+∫ex(x+1)2dx

∴T=ex−2(exx+1+∫ex(x+1)2dx−∫ex(x+1)2dx)=ex−2exx+1=ex(x+1)−2exx+1=ex(x−1)x+1

Step-by-step explanation: