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

1 Answer
Feb 25, 2018

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#