# Integration by Parts, 1/x dillema

• erjkism
In summary, the answer to \int\frac{1}{x}(dx) is lnx + c. but if you do it with integration by parts, you end up with \int\frac{1}{x}(dx) = 1 + \int\frac{1}{x}(dx)which comes to 0=1. why does this happen?The derivation of integration by parts comes from the product rule so the (uv) term is actually \int(uv)'. The constant from this integration is usually assumed to be combined with the constant from the other integral, but since the other integrals canceled out we still have a constant term 1 + c =
erjkism
the answer to $$\int\frac{1}{x}(dx)$$ is lnx +c. but if you do it with integration by parts, you end up with

$$\int\frac{1}{x}(dx)$$ = 1 + $$\int\frac{1}{x}(dx)$$

which comes to 0=1. why does this happen?

The derivation of integration by parts comes from the product rule so the (uv) term is actually $$\int(uv)'$$. The constant from this integration is usually assumed to be combined with the constant from the other integral, but since the other integrals canceled out we still have a constant term 1 + c = 0.

Though, this is just a guess, and I could be completely wrong.

I don't think it's possible to tell what you did wrong unless you give more details of your calculation. Did you start with something like this

$$\frac{1}{x} = \frac{1}{x} (Dx) = D\Big(\frac{1}{x}x\Big) - \Big(D\frac{1}{x}\Big)x = D\Big(\frac{1}{x}x\Big)+\frac{1}{x}?$$

But I'm not ending up into any paradox if I perform the integration by parts with this. You just get

$$\int_a^b \frac{1}{x}dx = \frac{b}{b} - \frac{a}{a} + \int_a^b\frac{1}{x}dx.$$

im not using a definite integral in my calculation. there aren't any limits "a" and "b"

Ah, but it's a trick! There is really no such thing as an indefinite integral. The notation

$$\int f(x) \; dx$$

is really just shorthand for

$$\int_a^x f(t) \; dt$$

for some arbitrary constant a. If you write it out like this, so that a is consistent, then you will resolve the paradox.

$$\int \left( \frac{d}{dx} f(x) \right) dx = f(x) + C$$

and you'll note that it is certainly true that

0 = 1 + C

for some C.

erjkism said:
im not using a definite integral in my calculation. there aren't any limits "a" and "b"

In a sense, the integral functions are definite integrals too. If F(x) is chosen so that its derivative is f(x), then it is

$$F(x) = \int_{x_0}^x f(u) du$$

with some $x_0$.

Your problem seems to be, that you are using a formula

$$\int fg' dx = fg - \int f'g dx$$

without knowing what it means. My advice is that try to understand how the formula is derived, and everything should get clearer.

It seems somebody was faster than me.

Last edited:

## 1. What is integration by parts?

Integration by parts is a method used in calculus to evaluate integrals that are in the form of a product of two functions. It involves finding a new integral by using the product rule of differentiation.

## 2. How do we apply integration by parts?

To apply integration by parts, we first identify the two functions in the integral and label one as u and the other as dv. Then, we use the product rule to find du and v, and substitute them into the formula ∫u dv = uv - ∫v du.

## 3. What is the purpose of using integration by parts?

The purpose of using integration by parts is to simplify an integral that is difficult to evaluate using other methods. It allows us to break down a complex integral into simpler parts, making it easier to solve.

## 4. How does the 1/x dilemma relate to integration by parts?

The 1/x dilemma refers to the difficulty in integrating functions that involve the natural logarithm of x. Integration by parts can be used to solve these types of integrals by choosing the natural logarithm as the dv term and using the formula ∫ln x dx = x(ln x - 1) + C.

## 5. What are some common mistakes to avoid when using integration by parts?

Some common mistakes to avoid when using integration by parts include selecting the wrong functions for u and dv, forgetting to use the product rule to find du and v, and not properly simplifying the resulting integral after applying the formula.

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