integrate it

 

[{  ( x^2 + 2x ) ^ 1/2  } + 3 ]^ -1

1/ ( (x²+2x) +3) dx

now put

x²+2x = t²

=> (x+1)² = t²+1

(2x+2)dx = 2tdt

(x+1)dx = tdt

so dx = tdt/(t²+1)

so given becomes

(tdt/ (t+3)((t²+1)

adding and subtracting 3 in the numerator

and splitting into 2 integrals

we get

dt/(t²+1) -3 dt/(t+3)(t²+1)

now in the first integral put t = tana

integral gets simplified to seca da

in the second also put t = tana

it becomes of the form

da/(sina+3cosa)

where we can put tana/2 = p

such that sina = 2p/(1+p²)     cosa = (1-p²)/(1+p²)    da = 2dp/(1+p²)

now this also is easy ..

after this question of back substituion

pls dont complain if the substitution takes more time than the working out part itself!!!