How do you integrate? #int(2x - 2)/(sqrtx +1) dx#

Question

#int(2x - 2)/(sqrtx +1) dx#

1410 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-07T17:04:57

#I=4/3x^(3/2)-2x+C#

We want to integrate

#I=int(2x-2)/(sqrt(x)+1)dx#

Rewrite the integrand

#I=2int(x-1)/(sqrt(x)+1)dx#

#color(white)(I)=2int((sqrt(x)+1)(sqrt(x)-1))/(sqrt(x)+1)dx#

#color(white)(I)=2intsqrt(x)-1dx#

Integrate by the power rule

#I=2(2/3x^(3/2)-x)+C#

#color(white)(I)=4/3x^(3/2)-2x+C#

Answer 2 · 2018-04-07T17:20:51

#I=4/3x^(3/2)-2x+c#

Here,

#I=int(2x-2)/(sqrtx+1)dx#

#=2int(x-1)/(sqrtx+1)dx#

#=2int[((sqrtx)^2-(1)^2)/(sqrtx+1)]dx#

#=2int((sqrtx-1)cancel((sqrtx+1)))/cancel((sqrtx+1))dx#

#=2int(sqrtx-1)dx#

#=2int(x^(1/2)-1)dx#

#=2[x^(3/2)/(3/2)-x]+c#

#=4/3x^(3/2)-2x+c#