What is the integral of x/(x-0.4)?

Question

Mathematica's Integrator found the answer to be

x + 0.4 ln(x-0.4) - 0.4

I'm assuming the answer's right, but I need to know how they got it. I've searched integral tables, but have not found anything similar and I couldn't integrate by parts (unless I did it wrong). I need the integral as part of a larger fluid mechanics problem, but I haven't done this type of integration in years.

Answer 1

y = x / (x - 2/5)

u = x - 2/5 ( and remember u + 2/5 = x)
du = 1 dx

y = (u + 2/5) / u
y = 1 + (2/5)/u

int( 1 + (2/5)/u )
= u + (2/5)*ln(u)
= (x-2/5) + (2/5)*ln(x-2/5)

Answer 2

x/(x-0.4)=1+0.4/(x-0.4) so integral = x+0.4 ln(x-0.4) +c where c is any arbitrary constant. Recollect integral 1/(x-a)=ln(x-a)+m which is an arbitrary constant again. Whenever you have the integrand as f(x)/g(x) where the numerator and denominator are polynomials, try splitting into partial fractions ensuring degree of numerator is lesser than that of denominator.

Answer 3

? 0 to ? 3tsin(5t)dt evaluate ? t sin(5t) dt combine by potential of things dv= sin(5t); v=-cos(5t) /5 u = t; du = dt; ? u dv = u v - ? v du ? t sin(5t) dt = (-a million/5) t cos(5t) - ? [-cos(5t) /5] dt = (-a million/5) t cos(5t) + (a million/5) ? cos(5t) dt = (-a million/5) t cos(5t) + (a million/5) sin(5t) /5 = (-a million/5) t cos(5t) + (a million/25) sin(5t) ?3 t sin(5t)dt = (-3/5) t cos(5t) + (3/25) sin(5t) permit F(t) = (-3/5) t cos(5t) + (3/25) sin(5t) F(?) = (-3/5) (?) cos(5?) + (3/25) sin(5?) = (-3/5)(?)(-a million) + (3/25)(0) = 3?/5 F(0) = (-3/5)(0) cos(0) + (3/25)sin(0) = 0 F(?) - F(0) = 3?/5 --------------------------------------... ? a million to 4 5(ln(x))/(x^2) dx evaluate ? ln (x) /x^2 dx combine by potential of things dv=a million/x^2 ; v=-a million/x; u = ln (x) ; du = a million/x dx ? u dv = u v - ? v du ? ln (x) /x^2 dx = - ln(x) / x + ? (a million/x) (a million/x) dx = - ln(x) / x + ? (a million/x^2) dx = -ln (x) / x - a million/x 5 ? ln (x) /x^2 dx = -5 ln(x) / x - 5/x permit F(x) = -5 ln(x) / x - 5/x F(4) = -5 ln(4) /4 - 5/4 = -2.9829 F(a million) = -5 ln(a million) /a million - 5/a million = 0 - 5 = -5 F(4)-F(a million) = -2.9829 - (-5) = 2.0171 ------------------------- word: d/dx (a million/x) = d/dx (x^(-a million)) = (-a million)x^(-2) = -a million/x^2 subsequently, crucial of -a million/x^2 = a million/x crucial of a million/x^2 = -a million/x --------------------------------------... ? a million to 4 ?(t) (ln(t))dt ? ?t ln(t) dt = ? t^(a million/2) ln(t) dt combine by potential of things dv= t^(a million/2) ; v= t^(a million/2+a million)/(a million/2+a million) ; v= (2/3) t^(3/2) u = ln(t); du = a million/t dt ? u dv = u v - ? v du ? t^(a million/2) ln(t) dt = (2/3) t^(3/2) ln(t) - (2/3) ? t^(3/2) /t dt = (2/3) t^(3/2) ln(t) - (2/3) ? t^(a million/2) dt = (2/3) t^(3/2) ln(t) - (2/3) [ t^(a million/2+a million)/(a million/2+a million)] = (2/3) t^(3/2) ln(t) - (2/3)(2/3) t^(3/2) = (2/3) t^(3/2) ln(t) - (4/9) t^(3/2) permit F(t) = (2/3) t^(3/2) ln(t) - (4/9) t^(3/2) F(4) = (2/3) 4^(3/2) ln(4) - (4/9) 4^(3/2) F(4) = (2/3)(8) ln(4) - (4/9) (8) = 3.8380 F(a million) = (2/3) a million^(3/2) ln(a million) - (4/9) a million^(3/2) F(a million) = 0 - 4/9 F(4)-F(a million) = 3.8380 - (-4/9) = 4.2824

Answer 4

∫dx / (ax + b) = (1/a) ln | ax + b | + c
∫xdx / (ax + b) = x/a - ( (b/a²) ln | ax + b | ) + c


so if ur equation is
dx / (x - 0.4)
then
∫dx/(x - 0.4 ) = ln | x - 0.4 | + c


BUT if the equation is
x dx / ( x - 0.4)
then
∫x dx / ( x - 0.4) = x + 0.4 ln| x - 0.4 | + c

Answer 5

Hi dear;

∫x/(x - 0.4) dx = -0.4 + (1*x )+ 0.4*Log[-0.4 + x] +c

Good Luck darling;

Answer 6

its simple...
In[x/(x-0.4)]....
Add and Subtract 0.4 in the Nr.
In{ (x-0.4+0.4) / (x -0.4) }
= In { [(x-0.4) / (x-0.4)] + [0.4 / (x-0.4)] } (i'm splitting the Nr into two fractions)
= In { [ 1 + [0.4 / (x-0.4)] }
= In (1) + In([0.4 / (x-0.4)] )
= x + 0.4*log(x-0.4) {we know integral of 1/(ax +b) is log (ax+b)/a
Adding the Integral Constant

IN (x/(x-0.4)) = x + 0.4*log(x-0.4) + c

Answer 7

it is x/(x-.4)
put x-.4 = t
x= t+.4
and dx/dt = 1
now f(x)dx/dt = ((t+.4)/t) = 1+.4/t
integarting wrt t we get t+.4 ln t = x-.4 +.4in(x-.4) = x+.4ln(x-.4) -.4

Answer 8

x/(x-0.4) = (x-0.4+0.4)/(x-0.4) = 1 + 0.4/(x-0.4).

Integrate to get,

x + 0.4 ln (x-0.4)