How do you evaluate the definite integral #int 2/sqrt(1+x)# from #[0,1]#? Calculus Introduction to Integration Definite and indefinite integrals 1 Answer Ratnaker Mehta Aug 8, 2016 #4(sqrt2-1)#. Explanation: Let #1+x=t^2 rArr dx=2tdt# Also, #x=0 rArr t^2=1 rArr t=1, &, x=1 rArr t=sqrt2#. Hence, #I=int_0^1 2/sqrt(1+x)dx=int_1^sqrt2 2/sqrt(t^2)*2tdt=4int_1^sqrt2 dt# #=4[t]_1^sqrt2# #=4(sqrt2-1)#. Answer link Related questions What is the difference between definite and indefinite integrals? What is the integral of #ln(7x)#? Is f(x)=x^3 the only possible antiderivative of f(x)=3x^2? If not, why not? How do you find the integral of #x^2-6x+5# from the interval [0,3]? What is a double integral? What is an iterated integral? How do you evaluate the integral #1/(sqrt(49-x^2))# from 0 to #7sqrt(3/2)#? How do you integrate #f(x)=intsin(e^t)dt# between 4 to #x^2#? How do you determine the indefinite integrals? How do you integrate #x^2sqrt(x^(4)+5)#? See all questions in Definite and indefinite integrals Impact of this question 1129 views around the world You can reuse this answer Creative Commons License