Question

friends I am preparing for nest help me in this question

find the integration of (2x - 1 )/ (4x^2 +4x + 2)^1/2

Answer 1

hello pari :-) 

we know that we can use partial fraction here :-) 

as we know that 2x - 1 = > A (8x+4) +B 

2x = > 8Ax 

we get A = >  1/4     and for B dear  we know that   2 *-1/2 -1  = > B 

mean B = > -2 

we got the values of A and B pari now we will find its integration 

let our function is I 

I =  ∫  (2x-1) dx /√(4x² +4x + 2)

we can write it is in form of {1/4 (8x + 4)- 2 }/√(4x² +4x + 2) 

now we will reduce this integration dear ! 

2×1/4√(4x² +4x + 2) dx  -  ∫2dx/√(4x² +4x + 2) 

which means => 1/2 √(2x +1)²+1 -  ∫2dx /√(4x² +4x + 2) 

we know that ∫ √x ² + a² = >  a²/2 ln ( x + √x ²+a² ) 

and let we solve I1 => /2 √(2x +1)²+1  and I2 => ∫-2dx/√(4x² +4x + 2)  and in last we will add both integration :-) its long but it will build over concept strong 

I1 = > 1/2 ∫√(2x +1)²+1 and here a = > 1  and x = > 2x +1)² 

I1 = > 1²/4 ln ( 2x+1 + √ (4x² +4x + 2) + C1  we got I1 

now we will solve I2 = > - ∫2dx/√(4x² +4x + 2)  now pari 

- ∫2dx/√ ( (2x +1 ) ² +1 ) 
 
and it is in the form  -2 ∫ dx/√ ( (2x +1 ) ² +1 )  actually now we know that 

∫ dx/ √a² + x² = >  log (x+ √ a² +x² ) and here a = >1  and x= > (2x+1) now putting values we get 

-2 ln ( 2x +1 √ (4x² +4x + 2 ) ) ×1/2   now we get  I2 = > -ln ( 2x +1 √ (4x² +4x + 2 ) ) 

now we know that I1 + I2 = > I adding both integration we get pari 

I = >  1/4 ln ( 2x+1 + √ (4x² +4x + 2) + C1 -  -ln ( 2x +1 √ (4x² +4x + 2 ) )  C2 

and c1 + c2 => constant so we named it C as a constant number 

1/4 ln ( 2x+1 + √ (4x² +4x + 2) +   -ln ( 2x +1 √ (4x² +4x + 2 ) )   + C is our required answer 

HOPES THIS WILL HELP YOU ! 

@ engineer gopal khandelwal ! IIT roorkey  btec :-)   

Answer 2

Hope this helps............