Find out the answer to "Solve the integral. ∫1udu"

aramutselv

Answered question

2021-12-10

Solve the integral.

\int \frac{1}{u} d u

Piosellisf

Beginner2021-12-11Added 40 answers

Step 1
Given:

\int \frac{1}{u} d u

Step 2
Explanation:
Common integral:

\int \frac{1}{u} d u = \int u^{- 1} d u = \ln | u | + C

Where

\frac{1}{a^{1}} = a^{- 1}

\int \frac{1}{u} d u = \ln | u | + C

Hence the solution.
Step 3
Consider,

y = \ln (x)

Therefore,

e^{y} = x

Differentiate above equation,

e^{y} d y = d x

Multiply both sides by

\frac{1}{e^{y} d x}

\frac{1}{e^{y} d x} \cdot e^{y} d y = \frac{1}{e^{y} d x} \cdot d x

Therefore,

\frac{d y}{d x} = \frac{1}{e^{y}}

Now put

x = e^{y}

\frac{d y}{d x} = \frac{1}{x}

y (x) = \ln (x)

So,

\frac{d y}{d x}

is the derivative of

\ln (x)

and it is

\frac{1}{x}

Since,

\frac{1}{x}

is the derivative of

\ln (x)

,
integral of

\frac{1}{x}

\ln (x)

by the fundamental theorem calculus
Hence the proof.

Jenny Bolton

Beginner2021-12-12Added 32 answers

Given:

\int \frac{1}{u} d u

This is the well-known tabular integral:

= \ln (u)

Applying a modulus to the logarithm argument expands its range:

\int \frac{1}{u} d u

Result:

= \ln (| u |) + C

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