Get the answer to "Evaluate the indefinite integral. ∫x+ln⁡xxdx"

High School
Calculus and Analysis
Calculus 2
Applications of integrals

eiraszero11cu

Answered question

2021-12-26

Evaluate the indefinite integral.

\int \frac{x + \ln x}{x} d x

Answer & Explanation

Corgnatiui

Beginner2021-12-27Added 35 answers

Step 1
The indefinite integral is given

\int \frac{x + \ln x}{x} d x

Simplify the indefinite integral.

\int \frac{x + \ln x}{x} d x = \int \frac{x}{x} d x + \int \frac{\ln x}{x} d x

= \int 1 \cdot d x + \int \frac{\ln x}{x} d x

= x + \int \frac{\ln x}{x} d x

Substitute

\ln x \Rightarrow

\frac{1}{x} d x \Rightarrow d t

Step 2
Apply the values integral,

\int \frac{\ln x}{x} d x = \int t d t

= \frac{t^{2}}{2} + c

.
Replace t with

\ln x

\int \frac{\ln x}{x} d x = \frac{{(\ln x)}^{2}}{2} + c

.
The final solution of the integral is,

\int \frac{x + \ln x}{x} d x = x + \frac{{(\ln x)}^{2}}{2} + C

Foreckije

Beginner2021-12-28Added 32 answers

\int \frac{\ln (x) + x}{x} d x

= \int (\frac{\ln (x)}{x} + 1) d x

= \int \frac{\ln (x)}{x} d x + \int 1 d x

\int \frac{\ln (x)}{x} d x

= \int u d u

= \frac{u^{2}}{2}

= \frac{\ln^{2} (x)}{2}

\int 1 d x

\int \frac{\ln (x)}{x} d x + \int 1 d x

= \frac{\ln^{2} (x)}{2} + x

\int \frac{\ln (x) + x}{x} d x

= \frac{\ln^{2} (x)}{2} + x + C

karton

Expert2022-01-04Added 613 answers

$\int \frac{x + \ln (x)}{x} d x$
Separate the fraction
$\int \frac{x}{x} + \frac{\ln (x)}{x} d x$
Divide
$\int 1 + \frac{\ln (x)}{x} d x$
Use properties of integrals
$\int 1 d x + \int \frac{\ln (x)}{x} d x$
Evaluate the integrals
$x + \frac{\ln (x)^{2}}{2}$
Add C
Answer:
$x + \frac{\ln (x)^{2}}{2} + C$

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