∫ √x / 2(1 + √x)² dx
let u = 1 + √x
u - 1 = √x
x = (u - 1)²
du = 1/2√x dx
= ∫ √x / 2u² *2√x du
= ∫ x / u² du
= ∫ (u - 1)² / u² du
= ∫ (u² - 2u + 1) / u² du
= ∫ (1 - 2/u + 1 / u²) du
= u -2ln(u) - 1/u + C
= (1 + √x) -2ln(1 + √x) - 1/(1 + √x) + K
= √x -2ln(1 + √x) - 1/(1 + √x) + C
(note here K + 1 = C)
you have u=1+sqrtx
du/dx=(1/2)x^(-1/2)
du=1/2)x^(-1/2)dx
THEN i'm pretty sure that you have to multiply by square root of x inside the integral and bring an x^2 outside the integral (its like multiplying by 1)
PS...DONT FORGET TO ADD +C !
hope this helps!
let 1+sqr(x)=t then dt= 1/2sqrt(x) *dx dx=2sqrt(x)dt
the integral will beintegral ( (t-1)^2/t^2)=integral((t^2-2t+1)/t^2
=integral( 1-2/t+1/t^2)=1-2lnt-1/t = 1-2ln(1+sqrtx)-1/(1+sqrx+1)
sqrt x / 2(1+ sqrt x )^2
= sqrt x / 2(1 + x)
= sqrt x / 2 + x
or
(x^1/2) / 2 + x
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Verified answer
∫ √x / 2(1 + √x)² dx
let u = 1 + √x
u - 1 = √x
x = (u - 1)²
du = 1/2√x dx
∫ √x / 2(1 + √x)² dx
= ∫ √x / 2u² *2√x du
= ∫ x / u² du
= ∫ (u - 1)² / u² du
= ∫ (u² - 2u + 1) / u² du
= ∫ (1 - 2/u + 1 / u²) du
= u -2ln(u) - 1/u + C
= (1 + √x) -2ln(1 + √x) - 1/(1 + √x) + K
= √x -2ln(1 + √x) - 1/(1 + √x) + C
(note here K + 1 = C)
you have u=1+sqrtx
du/dx=(1/2)x^(-1/2)
du=1/2)x^(-1/2)dx
THEN i'm pretty sure that you have to multiply by square root of x inside the integral and bring an x^2 outside the integral (its like multiplying by 1)
PS...DONT FORGET TO ADD +C !
hope this helps!
let 1+sqr(x)=t then dt= 1/2sqrt(x) *dx dx=2sqrt(x)dt
the integral will beintegral ( (t-1)^2/t^2)=integral((t^2-2t+1)/t^2
=integral( 1-2/t+1/t^2)=1-2lnt-1/t = 1-2ln(1+sqrtx)-1/(1+sqrx+1)
sqrt x / 2(1+ sqrt x )^2
= sqrt x / 2(1 + x)
= sqrt x / 2 + x
or
(x^1/2) / 2 + x