Use integration by parts to evaluate the following indefinite integrals.

(a) ∫ √x
ln xdx

(b)∫ x/cos^2 xdx

(c)∫ lnx/x^3 dx

(d)∫ ln(x^2 + 1)dx

Please show work, thanks!

∫ √(x)
ln(x) dx

u = ln(x). . . . dv = x^(1/2)
du = (1/x) dx . v = x^(1/2 + 1)/(1/2 + 1) ====> (2/3) * x^(3/2)

u * v - ∫ v * du

ln(x) * (2/3) * x^(3/2) - ∫ (2/3) * x^(3/2) * (1/x) dx

ln(x) * (2/3) * x^(3/2) - ∫ (2/3) * x^(2/2) dx

ln(x) * (2/3) * x^(3/2) - ∫ (2/3) * x^1 dx

ln(x) * (2/3) * x^(3/2) - (2/3) * x^2/2 + C

ln(x) * (2/3) * x^(3/2) - (1/3) * x^2 + C

===============

∫ [ x / cos^2(x) ] dx

∫ x * sec^2(x) dx

u = x. . . . . . .. dv = sec^2(x)
d = dx .. . . . . .v = tan(x)

u * v - ∫ v * du

x * tan(x) - ∫ tan(x) * dx

x * tan(x) - ∫ sin(x)/cos(x) * dx ===> u-sub as = cos(x) & du = -sin(x)

x * tan(x) + ∫ du/u

x * tan(x) + ln( u ) + C

x * tan(x) + lnI cos(x) I + C

==============

∫ ln(x) / x^3

u = ln(x). . . . dv = x^(-3)
du = (1/x) dx . v = x^(-3 + 1)/(-3 + 1) ====> (-1/2) * x^(-2)

u * v - ∫ v * du

ln(x) * (-1/2) * x^(-2) - ∫ (-1/2) * x^(-2) * (1/x) dx

ln(x) * (-1/2) * x^(-2) + ∫ (1/2) * (1/x^(2)) * (1/x) dx

ln(x) * (-1/2) * x^(-2) + ∫ (1/2) * (1/x^3) dx

ln(x) * (-1/2) * x^(-2) + ∫ (1/2) * x^-3 dx

ln(x) * (-1/2) * x^(-2) + (1/2) * x^(-3+1)/(-3+1) + C

ln(x) * (-1/2) * x^(-2) + (1/2) * x^(-2)/(-2) + C

ln(x) * (-1/2) * x^(-2) - (1/4) * x^(-2) + C

==============

∫ ln(x^2 + 1) dx

u = ln(x^2 + 1). . . . . . . . . dv = dx
du = (1/(x^2+1) ) * 2x dx . . . . . .v = x
du = 2x/(x^2+1)

u * v - ∫ v * du

ln(x^2 + 1) * x - ∫ x * 2x/(x^2+1) dx

ln(x^2 + 1) * x - ∫ 2x^2/(x^2+1) dx ===> long division

. . . . . 2
. . . . .. ____
x^2 + 1 I 2x^2
. . . . .-
. . . . . .2x^2 + 2
. . . . . ____________________
. . . . . . 0 - 2

2 - (2 / (x^2 + 1))


ln(x^2 + 1) * x - ∫ 2x^2/(x^2+1) dx ===> long division

ln(x^2 + 1) * x - ∫ [ 2 - (2 / (x^2 + 1)) ] dx

ln(x^2 + 1) * x - ∫ 2 + ∫ (2 / (x^2 + 1)) ] dx

ln(x^2 + 1) * x - 2x + 2tan^-1(x) + C

2/3 x^(3/2) lnx - 4/9 x^(3/2) , no idea , -(lnx)/2x^2-1/4x^2 , xln(x^2+1)-2x+2tan^-1 x