How do you use substitution to integrate x(sqrt(2x+1))dx.?
2 Answers
This is better-solved using integration by parts. There is nothing u-substitution-related that is easy to think of quickly, here.
I would choose the function that is easier to differentiate and make go away as
Let:
u = x du = dx dv = (2x+1)^(1//2)dx v = (2//3)/2(2x+1)^(3//2) = 1/3(2x+1)^(3//2)
Then, we get:
int udv = uv - intvdu
= x/3(2x+1)^(3//2) - int1/3(2x+1)^(3//2)dx
= x/3(2x+1)^(3//2) - 1/3(1/5(2x+1)^(5//2))
= x/3(2x+1)^(3//2) - 1/15(2x+1)^(5//2)
= (2x+1)^(3//2)[x/3 - 1/15(2x+1)]
= 1/15(2x+1)^(3//2)[5x - (2x+1)]
And now we tack on the constant at the end.
= color(blue)(ul(1/15(2x+1)^(3//2)(3x-1) + C))
Please see below.
Explanation:
Let
So
= 1/2 int 1/2(u-1)u^(1/2)du
= 1/4 int (u^(3/2)-u^(1/2)) du
= 1/4 (2/5u^(5/2) - 2/3u^(3/2)) +C
= 1/10u^(5/2)-1/6u^(3/2)+C
Simplify as desired before reversing the substitution.
I like
So we have
= sqrt(2x+1)/30 (2x+1) (6x-2) +C
= sqrt(2x+1)/15 (2x+1) (3x-1) +C