Help Finding Indefinite Integral ?

Oftentimes, with integration by parts, a single iteration isn't enough.  You'll have to do it over and over again until you reach an elementary integral or you can eliminate something

ln(x^3) * dx =>

3 * ln(x) * dx

int(ln(x) * dx)

u = ln(x) , du = dx / x , dv = dx , v = x

u * v - int(v * du) =>

x * ln(x) - int(x * dx/x) =>

x * ln(x) - int(dx) =>

x * ln(x) - x

3 * x * (ln(x) - 1) + C

int(e^(x) * sin(x) * dx)

u = sin(x) , du = cos(x) * dx , dv = e^(x) * dx , v = e^(x)

int(e^(x) * sin(x) * dx) = e^(x) * sin(x) - int(e^(x) * cos(x) * dx)

u = cos(x) , du = -sin(x) * dx , dv = e^(x) * dx , v = e^(x)

int(e^(x) * sin(x) * dx) = e^(x) * sin(x) - (e^(x) * cos(x) - int(-e^(x) * sin(x) * dx))

int(e^(x) * sin(x) * dx) = e^(x) * sin(x) - e^(x) * cos(x) + int(-e^(x) * sin(x) * dx)

int(e^(x) * sin(x) * dx) = e^(x) * (sin(x) - cos(x)) - int(e^(x) * sin(x) * dx)

Let int(e^(x) * sin(x) * dx) = k

k = e^(x) * (sin(x) - cos(x)) - k

2k = e^(x) * (sin(x) - cos(x))

k = (1/2) * e^(x) * (sin(x) - cos(x))

Add a constant of C

(1/2) * e^(x) * (sin(x) - cos(x)) + C

int(x^3 * sin(x) * dx)

In this case, using u = x^3 and proceeding from there will eventually get rid of the x component

u = x^3 , du = 3x^2 * dx , dv = sin(x) * dx , v = -cos(x)

-x^3 * cos(x) + 3 * int(x^2 * cos(x) * dx)

u = x^2 , du = 2x * dx , dv = cos(x) * dx , v = sin(x)

-x^3 * cos(x) + 3 * (x^2 * sin(x) - 2 * int(x * sin(x) * dx))

-x^3 * cos(x) + 3 * x^2 * sin(x) - 6 * int(x * sin(x) * dx)

u = x , du = dx , dv = sin(x) * dx , v = -cos(x)

-x^3 * cos(x) + 3 * x^2 * sin(x) - 6 * (-x * cos(x) + int(cos(x) * dx)) =>

-x^3 * cos(x) + 3 * x^2 * sin(x) + 6 * x * cos(x) - 6 * int(cos(x) * dx) =>

-x^3 * cos(x) + 3 * x^2 * sin(x) + 6 * x * cos(x) - 6 * sin(x) + C

int(arctan(x) * dx)

u = arctan(x)

du = dx / (1 + x^2)

dv = dx

v = x

x * arctan(x) - int(x * dx / (1 + x^2))

u = 1 + x^2

du = 2x * dx

x * arctan(x) - (1/2) * int(du / u) =>

x * arctan(x) - (1/2) * ln|u| + C =>

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

Here is one way to think of integrals of ln[f(x)] 

d/dx[ln(x^3)] = 3x^2/x^3 = 3/x 

d/dx[ln(x^3)*x] = ln(x^3)*1 + x*(3/x) = ln(x^3) + 3, so, 

∫ln(x^3)dx + ∫3dx = ln(x^3)*x 

∫ln(x^3)dx = xln(x^3) – 3x 

The next one you need to integrate by parts twice like this 

I = ∫s*e^x dx = s*e^x - ∫c*e^x dx = s*e^x -  c*e^x – I 

I = e^(x)[sin(x) -  cos(x)]/2 

Integrate by parts repeatedly, (just keep going), starting with 

J = ∫x^3*s dx = x^3*(-c) - ∫(-c)*3x^2 dx 

After gathering terms your result should be 

(6x – x^3)cos(x) + (3x^2 – 6)sin(x) 

One way to find K = ∫arctan(x)dx  relies on you already knowing  

d/dx[arctan(x)] = 1/(x^2 + 1) .....(if not see Note* below) 

d/dx[arctan(x)*x] = arctan(x) + x *1/(x^2 + 1) or  

arctan(x) = d/dx[arctan(x)*x] – (1/2)*2x/(x^2 + 1) and integrating 

∫arctan(x)dx = x*arctan(x) – (1/2)ln(x^2 + 1) 

Note: Sketch an acute triangle, angle y, opposite x adjacent 1. 

x/1 = tan(y) = sin(y)/cos(y)  ....use quotient rule 

dx/dy = [s*s – c*(-c)]/c^2, so,  

dy/dx = [cos(x)]^2 = 1/[√(x^2 + 1)]^2 = 1/(x^2 + 1)  from triangle sketch 

https://en.wikipedia.org/wiki/Antiderivative