Question

помогите решить интегрирование по частям int sqrtx lnx dx

Answer 1

∫ √x ln x dx =

√x = t →
x = t² → dx = 2t dt
∫ √x ln x dx = ∫ t ln (t²) (2t dt) =
∫ 2t² ln (t²) dt =
(t²) = u and 2t²dt = dv;
(2t / t²) dt = (2 / t) dt = du

 2(t²⁺¹)/(2+1) = (2/3)t³ = v


∫ 2t² ln (t²) dt = (2/3)t³ ln (t²) - ∫ (2/3)t³ (2/ t) dt =
(2/3)t³ ln (t²) - (4/3) ∫ (t³/ t) dt =
(2/3)t³ ln (t²) - (4/3) ∫ t² dt =
(2/3)t³ ln (t²) - (4/3)(t²⁺¹)/(2+1) + c =
(2/3)t³ ln (t²) - (4/3)(1/3)t³+ c =
(2/3)t³ ln (t²) - (4/9)t³+ c 
 t = √x
∫ √x ln x dx = (2/3)√x³ ln (√x²) - (4/9)√x³+ c = 
(2/3)x√x ln x - (4/9)x√x + c

Answer 2

$int{sqrt{x}lnx}, dx$

$U=lnx$

$dU=frac{1}{x}dx$

$V=frac{2}{3}*sqrt{x^3}$

$int{sqrt{x}lnx}, dx=frac{2}{3}*sqrt{x^3}lnx-int{frac{2}{3}sqrt{x^3}}, dx=frac{2}{3}*sqrt{x^3}lnx-int{frac{2}{3}sqrt{x^3}frac{1}{x}}, dx=frac{2}{3}*sqrt{x^3}lnx-frac{4}{9}sqrt{x^3}+C$