MathLab with Mr. Ahmed

MathLab with Mr. Ahmed is a math learning website by Ahmed Amasha, a mathematics teacher teaching the American curriculum. Explore Algebra 2, Precalculus, Calculus, and SAT and PSAT Math Prep through lessons, focused practice and printable resources.

MathLabwith Mr. Ahmed
Logarithmic, Exponential, and Other Transcendental Functions

5.2 The Natural Logarithmic Function: Integration

Focus on the reciprocal integral, logarithmic substitution, domain and absolute values.

Free preview Explore the lesson before creating an account.

Play games for this section

Before you solve

Ideas in this lesson

01The reciprocal integralWorked steps · five questions02Logarithmic substitutionWorked steps · five questions03Domain and absolute valuesWorked steps · five questions
Idea 1 of 3

The reciprocal integral

On a positive interval, ∫dxx=ln⁡x+C\int \frac{dx}{x} = \ln x + C. In general write ln⁡∣x∣\ln |x|.

Worked example 1

Evaluate ∫1e2x dx\int_{1}^{e} \dfrac{2}{x}\,dx.
  1. F(x)=2ln⁡xF(x) = 2 \ln x
  2. ln⁡(e)=1\ln (e) = 1; ln⁡1=0\ln 1 = 0
  3. 2[1−0]=22[1 - 0] = 2

Answer: 22

Worked example 2

Evaluate ∫1e23x dx\int_{1}^{e^{2}} \dfrac{3}{x}\,dx.
  1. F(x)=3ln⁡xF(x) = 3 \ln x
  2. ln⁡(e2)=2\ln (e^{2}) = 2; ln⁡1=0\ln 1 = 0
  3. 3[2−0]=63[2 - 0] = 6

Answer: 66

Idea 2 of 3

Logarithmic substitution

An integrand u′u\frac{u'}{u} has antiderivative ln⁡∣u∣\ln |u|. Check for a missing constant multiple.

Worked example 1

Evaluate ∫012xx2+2 dx\int_{0}^{1} \dfrac{2x}{x^{2} + 2}\,dx.
  1. u=x2+2u = x^{2} + 2; du=2xdxdu = 2x dx
  2. ∫231u du=ln⁡(3)−ln⁡(2)\int_{2}^{3} \dfrac{1}{u}\,du = \ln (3) - \ln (2)
  3. ln⁡(32)=ln⁡(32)≈0.41\ln (\dfrac{3}{2}) = \ln (\dfrac{3}{2}) \approx 0.41

Answer: ln⁡(32)≈0.41\ln (\dfrac{3}{2}) \approx 0.41

Worked example 2

Evaluate ∫011x+3 dx\int_{0}^{1} \dfrac{1}{x + 3}\,dx.
  1. u=x+3u = x + 3; du=1dxdu = 1 dx
  2. ∫341u du=ln⁡(4)−ln⁡(3)\int_{3}^{4} \dfrac{1}{u}\,du = \ln (4) - \ln (3)
  3. ln⁡(43)=ln⁡(43)≈0.29\ln (\dfrac{4}{3}) = \ln (\dfrac{4}{3}) \approx 0.29

Answer: ln⁡(43)≈0.29\ln (\dfrac{4}{3}) \approx 0.29

Idea 3 of 3

Domain and absolute values

Do not evaluate a definite logarithmic integral across a singularity as an ordinary endpoint subtraction.

Worked example 1

Evaluate ∫1e2x dx\int_{1}^{e} \dfrac{2}{x}\,dx.
  1. F(x)=2ln⁡xF(x) = 2 \ln x
  2. ln⁡(e)=1\ln (e) = 1; ln⁡1=0\ln 1 = 0
  3. 2[1−0]=22[1 - 0] = 2

Answer: 22

Worked example 2

Evaluate ∫1e23x dx\int_{1}^{e^{2}} \dfrac{3}{x}\,dx.
  1. F(x)=3ln⁡xF(x) = 3 \ln x
  2. ln⁡(e2)=2\ln (e^{2}) = 2; ln⁡1=0\ln 1 = 0
  3. 3[2−0]=63[2 - 0] = 6

Answer: 66

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §5.2,p5.2, p. 328328. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.