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```
f(x) is differentiable function satisfying f(x).f’(-x)=f(-x).f’(x), and f(0)=1, find value of integration of 1/{1+f(x)} from -5 to 5

```
2 years ago Deepak Kumar Shringi
4403 Points
```							here you can say f(x) is a even function let f(x)=x^2+1 now when you integrate an even function you wil get odd fucntion so from -5 to 5 limit it will be zero
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2 years ago
```							if f(x) is even function, integration of f(x) from -5 to 5 = 2. integration of f(x) from 0 to 5, so how I get 0? if f(x) is an odd function, the result will be 0.
```
2 years ago Deepak Kumar Shringi
4403 Points
```							?
```
2 years ago Deepak Kumar Shringi
4403 Points
```							please check this is a very basic thing that if f(x) is even its integration will be odd you asked integration of 1/1+f(x) not about f(x)
```
2 years ago
```							let f(x)=x^2 -1. 1/1+f(x) =1/x^2 is even also. now integrating from -5 to 5 how I get 0? plz clarify.
```
2 years ago Deepak Kumar Shringi
4403 Points
```							?
```
2 years ago Deepak Kumar Shringi
4403 Points
```							you have choosen wrong function f(0) should be 1
```
2 years ago
```							sorry. but take g(x)=1/1+f(x)g(-x)=g(x). so g(x) is even. now integration of g(x) from-5 to5 = 2*integration of g(x) from 0 to 5
```
2 years ago
```							I tried and got the result.f(x).f’(-x)=f(-x).f’(x)i.e. f’(x)/f(x)=f’(-x)/f(-x)integrating and using f(0)=1, we get f(x).f(-x)=1...(1)integration of 1/1+f(x) from -5 to 5=integration from 0 to 5 of [1/1+f(x) +1/1+f(-x)]=integration of {f(x)+1}/{f(x)+1} from 0 to 5 [ using (1)]=5
```
2 years ago
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