Asked by Sara
Find the integral of:
2xsqrt(x+1) dx
2xsqrt(x+1) dx
Answers
Answered by
Patricia
Use wolframalpha
Saved me.
Saved me.
Answered by
Reiny
let u = 2x
du = 2 dx
let dv = (x+1)^(1/2) dx
v = (2/3)(x+1)^(3/2)
∫2xsqrt(x+1) dx = uv - ∫v du
= (2x(2/3)(x+1)^(3/2) - ∫(4/3)(x+1)^(3/2) dx
= (4/3)x(x+1)^(3/2) - (8/15)(x+1)^(5/2)
= (1/15)(x+1)^(3/2) (20x - 8(x+1) )
= (4/15)(x+1)^(3/2) (5x - 2x - 2)
= (4/15)(x+1)^(3/2) (3x - 2)
du = 2 dx
let dv = (x+1)^(1/2) dx
v = (2/3)(x+1)^(3/2)
∫2xsqrt(x+1) dx = uv - ∫v du
= (2x(2/3)(x+1)^(3/2) - ∫(4/3)(x+1)^(3/2) dx
= (4/3)x(x+1)^(3/2) - (8/15)(x+1)^(5/2)
= (1/15)(x+1)^(3/2) (20x - 8(x+1) )
= (4/15)(x+1)^(3/2) (5x - 2x - 2)
= (4/15)(x+1)^(3/2) (3x - 2)
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