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# Intergral Partial Fraction

With the previous method we cannot (will be difficult) to neutralize the forms as below
∫ 5x (2x – 4) dx or ∫ 2x sin x dx.

To solve the form of the equation, let us consider the following steps.
If f and g are differential functions, the rule applies.

If we integrate the rule, we will get it

To further simplify the form we can write it with

u = f(x)            du = f ’(x) dx
v = g(x)           dv = g ‘(x) dx

so, the formula can be simplified as follows.

∫ u dv = uv - ∫ v du

Example 1.
Solve the integral results of ∫ 12x (2x - 3)3 dx
If it is known that,

So that,

Example 2.
Solve the integral results of ∫ 12x (2x - 3)3 dx
If it is known that,

So that,

### 1 Response to "Intergral Partial Fraction"

1. When fractions are being divided, as a fraction you need to "flip" the second fraction and change the operation sign from division to multiplication. 7/11 now becomes 11/7. You will now multiply the fractions.