𝛗:=∫_0 ^( 1) (√x) ln((1/(1−x)))dx solution: 𝛗:= ∫_0 ^( 1) (√x) Σ_(n=1) ^∞ (x^n /n) dx :=Σ_(n=1) ^∞ (1/n)∫_0 ^( 1) x^(n+(1/2)) dx := Σ_(n=1) ^∞ (1/(n(n+(3/2)))) = (2/3)Σ_(n=1) ^∞ ((1/n)−(1/(n+(3/2)))) :=(2/3){γ −γ+Σ_(n=1) ^∞ ((1/n)−(1/(n+(3/2)))) } := (2/3) γ +(2/3) ψ(1+(3/2))=(2/3) γ +(2/3)((2/3)+ψ((3/2))) :=(2/3) γ +(4/9) +(2/3)(2+ψ((1/2))) :=(2/3) γ +((16)/9) −(2/3) γ−(4/3) ln(2) :=((16)/9) −(4/3) ln(2) .....✓


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Question Number 139811 by mnjuly1970 last updated on 01/May/21
$𝛗:=∫_0 ^( 1) (√x) ln((1/(1−x)))dx solution: 𝛗:= ∫_0 ^( 1) (√x) Σ_(n=1) ^∞ (x^n /n) dx :=Σ_(n=1) ^∞ (1/n)∫_0 ^( 1) x^(n+(1/2)) dx := Σ_(n=1) ^∞ (1/(n(n+(3/2)))) = (2/3)Σ_(n=1) ^∞ ((1/n)−(1/(n+(3/2)))) :=(2/3){γ −γ+Σ_(n=1) ^∞ ((1/n)−(1/(n+(3/2)))) } := (2/3) γ +(2/3) ψ(1+(3/2))=(2/3) γ +(2/3)((2/3)+ψ((3/2))) :=(2/3) γ +(4/9) +(2/3)(2+ψ((1/2))) :=(2/3) γ +((16)/9) −(2/3) γ−(4/3) ln(2) :=((16)/9) −(4/3) ln(2) .....✓$
$ϕ := \int_{0}^{1} \sqrt{x} l n (\frac{1}{1 - x}) d x$ $s o l u t i o n :$ $ϕ := \int_{0}^{1} \sqrt{x} \underset{n = 1}{\sum^{\infty}} \frac{x^{n}}{n} d x$ $:= \underset{n = 1}{\sum^{\infty}} \frac{1}{n} \int_{0}^{1} x^{n + \frac{1}{2}} d x$ $:= \underset{n = 1}{\sum^{\infty}} \frac{1}{n (n + \frac{3}{2})} = \frac{2}{3} \underset{n = 1}{\sum^{\infty}} (\frac{1}{n} - \frac{1}{n + \frac{3}{2}})$ $:= \frac{2}{3} {γ - γ + \underset{n = 1}{\sum^{\infty}} (\frac{1}{n} - \frac{1}{n + \frac{3}{2}})}$ $:= \frac{2}{3} γ + \frac{2}{3} ψ (1 + \frac{3}{2}) = \frac{2}{3} γ + \frac{2}{3} (\frac{2}{3} + ψ (\frac{3}{2}))$ $:= \frac{2}{3} γ + \frac{4}{9} + \frac{2}{3} (2 + ψ (\frac{1}{2}))$ $:= \frac{2}{3} γ + \frac{16}{9} - \frac{2}{3} γ - \frac{4}{3} l n (2)$ $:= \frac{16}{9} - \frac{4}{3} l n (2) . . . . . ✓$
	Answered by mathmax by abdo last updated on 01/May/21
	$Φ=∫_0 ^1 (√x)log((1/(1−x)))dx ⇒Φ=−∫_0 ^1 (√x)log(1−x)dx =_((√x)=t) −∫_0 ^1 t log(1−t^2 )(2t)dt =−2∫_0 ^1 t^2 log(1−t^2 )dt =−2{ [((t^3 −1)/3)log(1−t^2 )]_0 ^1 −∫_0 ^1 ((t^3 −1)/3)×((−2t)/(1−t^2 ))dt} =−(4/3) ∫_0 ^1 ((t^4 −t)/(1−t^2 ))dt =(4/3)∫_0 ^1 ((t(t^3 −1))/((t−1)(t+1)))dt =(4/3) ∫_0 ^1 ((t(t^2 +t+1))/(t+1))dt =(4/3)∫_0 ^1 ((t^3 +t^2 +t)/(t+1))dt =_(t+1=y) (4/3)∫_1 ^2 (((y−1)^3 +(y−1)^2 +y−1)/y)dy =(4/3)∫_1 ^2 ((y^3 −3y^2 +3y−1+y^2 −2y+1+y−1)/y)dy =(4/3)∫_1 ^2 ((y^3 −2y^2 +2y−1)/y)dy =(4/3)∫_1 ^2 (y^2 −2y+2−(1/y))dy =(4/3)[(y^3 /3)−y^2 +2y−log∣y∣]_1 ^2 =(4/3){(8/3)−4+4−log2−(1/3)+1−2} =(4/3){(7/3)−1−log2} =(4/3)((4/3)−log2) =((16)/9)−(4/3)log2$
	$Φ = \int_{0}^{1} \sqrt{x} \log (\frac{1}{1 - x}) dx \Rightarrow Φ = - \int_{0}^{1} \sqrt{x} \log (1 - x) dx$ $=_{\sqrt{x} = t} - \int_{0}^{1} t \log (1 - t^{2}) (2 t) dt$ $= - 2 \int_{0}^{1} t^{2} \log (1 - t^{2}) dt$ $= - 2 {{[\frac{t^{3} - 1}{3} \log (1 - t^{2})]}_{0}^{1} - \int_{0}^{1} \frac{t^{3} - 1}{3} \times \frac{- 2 t}{1 - t^{2}} dt}$ $= - \frac{4}{3} \int_{0}^{1} \frac{t^{4} - t}{1 - t^{2}} dt = \frac{4}{3} \int_{0}^{1} \frac{t (t^{3} - 1)}{(t - 1) (t + 1)} dt$ $= \frac{4}{3} \int_{0}^{1} \frac{t (t^{2} + t + 1)}{t + 1} dt = \frac{4}{3} \int_{0}^{1} \frac{t^{3} + t^{2} + t}{t + 1} dt$ $=_{t + 1 = y} \frac{4}{3} \int_{1}^{2} \frac{{(y - 1)}^{3} + {(y - 1)}^{2} + y - 1}{y} dy$ $= \frac{4}{3} \int_{1}^{2} \frac{y^{3} - {3 y}^{2} + 3 y - 1 + y^{2} - 2 y + 1 + y - 1}{y} dy$ $= \frac{4}{3} \int_{1}^{2} \frac{y^{3} - {2 y}^{2} + 2 y - 1}{y} dy$ $= \frac{4}{3} \int_{1}^{2} (y^{2} - 2 y + 2 - \frac{1}{y}) dy$ $= \frac{4}{3} {[\frac{y^{3}}{3} - y^{2} + 2 y - \log ∣ y ∣]}_{1}^{2}$ $= \frac{4}{3} {\frac{8}{3} - 4 + 4 - \log 2 - \frac{1}{3} + 1 - 2}$ $= \frac{4}{3} {\frac{7}{3} - 1 - \log 2} = \frac{4}{3} (\frac{4}{3} - \log 2) = \frac{16}{9} - \frac{4}{3} \log 2$
	Commented by mnjuly1970 last updated on 01/May/21

	$t h x s i r m a x . . .$
	Commented by mathmax by abdo last updated on 02/May/21

	$you are welcome sir$
	Answered by Ar Brandon last updated on 16/May/21
	$φ=∫_0 ^1 (√x)ln((1/(1−x)))dx=−∫_0 ^1 (√x)ln(1−x)dx ∫_0 ^1 x^(m−1) (1−x)^(n−1) dx=β(m,n)=((Γ(m)Γ(n))/(Γ(m+n))) lnβ(m,n)=lnΓ(m)+lnΓ(n)−lnΓ(m+n) (1/(β(m,n)))∙((∂β(m,n))/∂n)=ψ(n)−ψ(m+n) ⇒((∂β((3/2),1))/∂n)=β((3/2),1){ψ(1)−ψ((5/2))} =((Γ((3/2)))/(Γ((5/2)))){−γ−((2/3)+2−γ−2ln2)} φ=−(2/3)(2ln2−(8/3))=((16)/9)−((4ln2)/3)$
	$ϕ = \int_{0}^{1} \sqrt{x} \ln (\frac{1}{1 - x}) dx = - \int_{0}^{1} \sqrt{x} \ln (1 - x) dx$ $\int_{0}^{1} x^{m - 1} {(1 - x)}^{n - 1} dx = β (m, n) = \frac{Γ (m) Γ (n)}{Γ (m + n)}$ $\ln β (m, n) = \ln Γ (m) + \ln Γ (n) - \ln Γ (m + n)$ $\frac{1}{β (m, n)} \cdot \frac{\partial β (m, n)}{\partial n} = ψ (n) - ψ (m + n)$ $\Rightarrow \frac{\partial β (\frac{3}{2}, 1)}{\partial n} = β (\frac{3}{2}, 1) {ψ (1) - ψ (\frac{5}{2})}$ $= \frac{Γ (\frac{3}{2})}{Γ (\frac{5}{2})} {- γ - (\frac{2}{3} + 2 - γ - 2 \ln 2)}$ $ϕ = - \frac{2}{3} (2 \ln 2 - \frac{8}{3}) = \frac{16}{9} - \frac{4 \ln 2}{3}$
	Commented by mnjuly1970 last updated on 01/May/21

	$g r a t e f u l m r b r a n d o n . . .$
	Commented by SLVR last updated on 01/May/21

	$c o u l d y o u e x p l a i n w h a t ψ (1) ? ψ (5 / 2)$ $T h a n k s m r . B r a n d o n s i r$
	Commented by Ar Brandon last updated on 01/May/21

	${You}^{'} re welcome Sir$
	Commented by Ar Brandon last updated on 01/May/21

	$Γ^{'} (x) = Γ (x) ψ (x) \Rightarrow \frac{Γ^{'} (x)}{Γ (x)} = ψ (x)$ $ψ (s + 1) = \frac{1}{s} + ψ (s), ψ (\frac{1}{2}) = - γ - 2 \ln 2$ $ψ (x) = - γ + \underset{n = 0}{\sum^{\infty}} (\frac{1}{n + 1} - \frac{1}{n + x}) \Rightarrow ψ (1) = - γ$ $ψ (x) represents digamma function .$
	Commented by SLVR last updated on 01/May/21

	$S o . . . . k i n d y o u . . s i r G r e a t . . . e x p l a n a t i o n$ $G o d . . . b l e s s y o u f o r e v e r$
	Commented by Ar Brandon last updated on 01/May/21

	$Thanks!$
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