How do you integrate int x/sqrt(4x^2+4x+-24)dx using trigonometric substitution?

1 Answer
Aug 5, 2018

I=1/2sqrt(x^2+x+-6)-1/4ln|x+1/2+sqrt(x^2+x+-6)|+C

Explanation:

Without trigonometric substitution :

Here ,

I=intx/sqrt(4x^2+4x+-24)dx

=1/2intx/sqrt(x^2+x+-6)dx

=1/4int(2x)/sqrt(x^2+x+-6)dx

=1/4int(2x+1-1)/sqrt(x^2+x+-6)dx

=1/4int(2x+1)/sqrt(x^2+x+-6)dx-1/4int1/sqrt(x^2+x+-6)dx

=1/4int(d(x^2+x+-6))/sqrt(x^2+x+-6)-1/4I_1

=1/4*2sqrt(x^2+x+-6)-1/4I_1

=1/2sqrt(x^2+x+-6)-1/4I_1.....................to(A)

Now ,

I_1=int1/sqrt(x^2+x+-6)dx

Now ,

(i)x^2+x+6=x^2+x+1/4+23/4=(x+1/2)^2+23/4

(ii)x^2+x-6=x^2+x+1/4-25/4=(x+1/2)^2-25/4

So ,

I_1=int1/sqrt((x+1/2)^2+K)dx ,where ,K=23/4 orK=-25/4

:.I_1=ln|(x+1/2)+sqrt((x+1/2)^2+K)|+c'

:.I_1=ln|(x+1/2)+sqrt(x^2+x+1/4+K)|+c'

substitute K=23/4 orK=-25/4

:.I_1=ln|x+1/2+sqrt(x^2+x+-6)|+c'

From (A) we get

I=1/2sqrt(x^2+x+-6)-1/4ln|x+1/2+sqrt(x^2+x+-6)|+C

Note :

color(red)(int(f'(x))/sqrt(f(x))dx=2sqrt(f(x))+c

OR

color(red)(int(d[f(x)])/sqrt(f(x))=2sqrt(f(x))+c