IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

sankydreams

integrate [(a + x) / x]^1/2

thedumbheadwithnobrain

(a+x)^1/2*(x)^1/2 + a/2[log((a+x)^1/2 + (x)^1/2)] +C

ramyadiamond

U should put x=atan²y, then the root will be removed. Try solving it afterwards.

coool_shetty

Excellent answer ramya!!!!

thedumbheadwithnobrain

answer is 2a[(x/a)^1/2*(1+x/a)^1/2 + 1/2 log((x/a)^1/2 + (1+x/a)^1/2)] + C

neeraj_agarwal_1990

x^(-1/2) . (a+x)^(1/2)

(m+1)/n +p =1 is an integer....
so take out x common frm a+x and put 1+a/x = t^2

chotasoniji

X=A TAN*TANX

priyesh

here is the simplest method without any unusual substitution like atan² y etc

((a+x)/x)

multiply num & denominator by

(a + x)

= (a+x) /

(x² + ax)

multiply divide by 2

1/2 [ (2x + a + a) /

(x² + ax) ]

1/2 [ (2x + a) /

(x² + ax) + a /

(x² + ax) ]

integrate the two terms in square bracket indivisually

for (2x + a) /

(x² + ax) put x² + ax = t you will get numerator as dt so u will get dt/

t

for a /

(x² + ax) write x² + ax as (x + a/2)² - (a/2)² & use the standard result of 1/

x² - a²

hope you understood

cheers