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Calculus Level 5

Consider the integral, $I=\displaystyle \int \dfrac{1}{x} dx$

Step 1: $\displaystyle \int \dfrac{1}{x}(1) dx =\dfrac{1}{x}x-\int \bigg( - \dfrac{1}{x^2} \bigg ) xdx$

Step 2: $\displaystyle \int \dfrac{1}{x} dx=1+\int \dfrac{1}{x} dx$

Step 3: $\displaystyle \int \dfrac{1}{x} dx-\int \dfrac{1}{x} dx=1$

Step 4: $0=1$

Which of the above step/s is/are wrong?

If steps a,b,c... are wrong, enter your answer as $a \times b \times c \times ..$

1 solution

Zach Abueg
Jan 15, 2017

Indefinite integrals always include a constant of integration $C$ .

$\displaystyle \int \dfrac {1}{x} dx = ln$ $x + C$

$\displaystyle \int \dfrac {1}{x} dx - \int \dfrac {1}{x} dx = (ln$ $x + C) - (ln$ $x + C) = C_1 - C_2$

Because the two $C$ 's are not necessarily equal, we cannot conclude that $C_1 - C_2 = 0$ .

Thus, step $4$ is incorrect.

in step 2,

are you saying $lnx+C_{1}=lnx+C_{2}+1$ can be concluded

Rohith M.Athreya - 4 years, 4 months ago

That's a feasible statement, yes.

Zach Abueg - 4 years, 4 months ago

but arent the two arbitrary constants?

Rohith M.Athreya - 4 years, 4 months ago

@Rohith M.Athreya – Yes, we do not know them and technically the $+ 1$ is part of $C_2$ . Perhaps I don't understand your question.

Zach Abueg - 4 years, 4 months ago

Can anybody explain me what is the significance of $C_1-C_2=1$ ?

Kalpok Guha - 4 years ago