#### To determine

**To find:**

An explicit formula for fx

#### Answer

fx=xcosx+sinx1+x2x2

#### Explanation

**1) Concept:**

TheFirst FundamentalTheorem for calculus: If f is continuous on [a, b], then the function g is defined by g(x)=∫abftdta≤x≤b is continuous on [a,b] and differentiable on (a,b) and g’(x)=f(x)

**2) Calculation:**

The given integral is

∫0xftdt=xsinx+∫0xf(t)1+t2dt

Now by first fundamental theorem of calculus, the given function f is continuous

Therefore, take derivative of both sides with respect to x

ddx∫0xftdt=ddxxsinx+ddx∫0xf(t)1+t2dt

Here all the terms are functions of x

Therefore, the equation becomes

fx=xcosx+sinx+fx1+x2

Combine the terms of fx

fx-fx1+x2=xcosx+sinx

fx1-11+x2=xcosx+sinx

fxx21+x2=xcosx+sinx So

fx=xcosx+sinx1+x2x2

**Conclusion:**

An explicit formula for fx=xcosx+sinx1+x2x2