

The idea is to split the integrand into two parts. One part you'll call u, and the other you'll call dv, so that the product of u and dv is the original integrand. Then, from u you will obtain du by differentiating, and from dv you'll obtain v by integrating. The trick here is to let dv be something that is not only easy to integrate but doesn't get "worse" when you integrate it, and u is the rest of the equation  something that gets "better" when you differentiate it.
You see, dv will need to be integrated twice, so it's best to pick things like e^{x} dx, sin(x) dx, or (x+a)^{n} dx as dv, because you can integrate these twice without too much pain. Meanwhile, if you pick u as something that gets nicer  such as disappearing  when you differentiate it, then the scheme will work. For example, if you let u=x then du=dx, which is helpful because the x more or less disappears.
The "ILATE" method is good for picking dv so that it is easy to integrate, leaving you with something "better". The idea is to let dv be something as low in the ILATE table as possible. ILATE stands for:
Function Category (click category to jump to that table) I: inverse trigonometric functions: arctan(x), arcsec(x), etc. L: logarithmic functions: ln(x), log_{2}(x), etc. A: algebraic functions: x^{2}, 3x^{50}, etc. T: trigonometric functions: sin(x), tan(x), etc. E: exponential functions: e^{x}, 13^{x}, etc.
The logical two halves of this are x^{4} and ln(x). In the ILATE table, x^{4} is "A" for algebraic, and ln(x) is "L" for logarithmic. x^{4} comes later in the table, so we'll make dv=x^{4} dx, which leaves u=ln(x).∫x^{4} ln(x) dx
dv = x^{4} dx du = (1/x) dx v = (1/5) x^{5} u = ln(x)
∫u dv = uv  ∫v du
∫(ln x)(x^{4}dx) = (ln x)(1/5)x^{5}  ∫(1/5)x^{5}(1/x)dx
= (ln x)(1/5)x^{5}  ∫(1/5)x^{4}dx
= (1/5)(ln x)x^{5}  (1/25)x^{5}
Here, we let dv = sqrt(x+a) dx, which leaves u=x, effectively making the "x" disappear. sqrt(x+a) is (x+a)^{1/2}, which can be integrated twice fairly easily.∫x sqrt(x+a) dx
dv = (x+a)^{1/2} du = dx v = (2/3)(x+a)^{3/2} u = x
∫u dv = uv  ∫v du
∫x(x+a)^{1/2} dx = (2/3)(x)(x+a)^{3/2}
 ∫(2/3)(x+a)^{3/2}dx
= (2/3)(x)(x+a)^{3/2}(4/15)(x+a)^{5/2}
= ((2/3)(x)(4/15)(x+a))(x+a)^{3/2}
= (2/15)(3x2a)(x+a)^{3/2}
This example generalizes the previous example, taking any power of (x+a):
Here, we let dv = (x+a)^{k} dx, which leaves, as before, u=x, making the "x" disappear again. (x+a)^{k} can be integrated twice fairly easily.∫x (x+a)^{k} dx
dv = (x+a)^{k} du = dx v = (1/(k+1))(x+a)^{k+1} u = x
∫u dv = uv  ∫v du
∫x(x+a)^{k} dx = (1/(k+1))(x)(x+a)^{k+1}
 ∫(1/(k+1))(x+a)^{k+1}dx
= (1/(k+1))(x)(x+a)^{k+1}(1/((k+1)(k+2)))(x+a)^{k+2}
= (x/(k+1)(x+a)/((k+1)(k+2)))(x+a)^{k+1}
= (x(k+2)(x+a))/((k+1)(k+2))(x+a)^{k+1}
= ((k+1)xa)/((k+1)(k+2))(x+a)^{k+1}
. . . . . . a graphical representation will be added to this web page. To make it clear, we will express the integration by parts formula as
∫u dv + ∫v du = uv,
and treat u and v as functions of an independent parameter, t. Now, plot the locus of (u,v) on axes, (u horizontal, v vertical), and it becomes clear that the area, uv, is composed of the sum of the two integrals, as each one views the area "under" the curve as being between the curve and its axis. That is, ∫v du represents the area between the curve and the u axis, while ∫u dv represents the area between the curve and the v axis.
Wikipedia: Integration by parts
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