Question

How do you integrate `int_0^1sqrt(1+1/(4x))dx` ?

Answer 1

The answer is `=1.48`

Explanation:

First calculate the indefinite integral

`sqrt(1+1/4x)=sqrt(1/4(4+1/x))=1/2sqrt(4+1/x)`

Perform the substitution

`u=sqrt(4+1/x)`,

`u^2=4+1/x`, `=>`, `1/x=u^2-4`

`du=1/2(4+1/x)^(-1/2)*-1/x^2dx`

`=-1/2*1/u*(u^2-4)^2dx`

Therefore,

`intsqrt(1+1/(4x))=1/2intsqrt(4+1/x)dx`

`=-int(u^2*du)/(u^2-4)^2`

`=-int((u^2-4+4)du)/((u^2-4)^2)`

`=-int(du)/(u^2-4)-int(4du)/(u^2-4)^2`

Perform the decomposition into partial fractions on the first integral

`1/(u^2-4)=1/((u+2)(u-2))=A/(u+2)+B/(u-2)`

`=(A(u-2)+B(u+2))/((u+2)(u-2))`

Comparing the numerators

`1=A(u-2)+B(u+2)`

Let `u=-2`, `=>`, `A=-1/4`

Let `u=2`, `=>`, `B=1/4`

Therefore,

`-int(du)/(u^2-4)=1/4int(du)/(u+2)-1/4int(du)/(u-2)`

`=1/4ln(u+2)-1/4ln(u-2)`

Perform the decomposition into partial fractions on the second integral

`1/(u^2-4)^2=1/((u+2)^2(u-2)^2)`

`=A/(u+2)+B/(u+2)^2+C/(u-2)+D/(u-2)^2`

`=(A(u+2)(u-2)^2+B(u-2)^2+C(u+2)^2(u-2)+D(u+2)^2)/((u+2)^2(u-2)^2)`

Compare the numerators

`1=(A(u+2)(u-2)^2+B(u-2)^2+C(u+2)^2(u-2)+D(u+2)^2)`

Let `u=2`, `=>`, `D=1/16`

Let `u=-2`, `=>`, `B=1/16`

Coefficients of `u^3`

`0=A+C`, `=>`, `A=-C`

Let `u=0`, `=>` `1=8A+4B-8C+4D`

`8(A-C)=1-1/2=1/2`

`A-C=1/16`

`A=1/32`

`C=-1/32`

Therefore,

`-int(4du)/(u^2-4)^2=-4int((1/32)/(u+2)+(1/16)/(u+2)^2+(-1/32)/(u-2)+(1/16)/(u-2)^2)du`

`=-1/8ln(u+2)+1/4*1/(u+2)+1/8ln(u-2)+1/4*1/(u-2)`

Putting all together

`intsqrt(1+1/(4x))=1/4ln(u+2)-1/4ln(u-2)-1/8ln(u+2)+1/4*1/(u+2)+1/8ln(u-2)+1/4*1/(u-2)`

`int_0^1sqrt(1+1/(4x))=[1/4ln(u+2)-1/4ln(u-2)-1/8ln(u+2)+1/4*1/(u+2)+1/8ln(u-2)+1/4*1/(u-2)]_2^sqrt5`

`=[1/8ln(u+2)-1/8ln(u-2)+1/(4(u+2))+1/(4(u-2))]_2^sqrt5`

`=1.48`