Question

How do you evaluate `\int _ { 0} ^ { 1} \frac { x d x } { \sqrt { x ^ { 2} + 1} }`?

Answer 1

`int_0^1(xdx)/sqrt(x^2+1)=sqrt2-1`

Explanation:

`I=int_0^1(xdx)/sqrt(x^2+1)`

We will use the substitution `u=x^2+1=>du=2xdx`. We will need to multiply the integrand by `2`. Once we do this, then we can substitute the variables.

When we substitute, we will also have to change the bounds using `u=x^2+1` as follows:

• `x=0" "=>" "u=0^2+1=1`
• `x=1" "=>" "u=1^2+1=2`

So we get:

`I=1/2int_0^1(2xdx)/sqrt(x^2+1)=1/2int_1^2(du)/sqrtu=1/2int_1^2u^(-1/2)du`

Now using `intu^ndu=u^(n+1)/(n+1)+C`:

`I=1/2[u^(1/2)/(1/2)]_1^2=[sqrtu]_1^2=sqrt2-1`