Math Problem Statement

Solve the following integrals: 1. ∫ 1/(1-x) dx, 2. ∫ 5/(2x+7) dx, 3. ∫(from -1 to 2) 2x/(x^2+4) dx, 4. ∫(from 0 to ln 2) e^(-3x) dx, 5. ∫ e^(1/x)/x^2 dx, 6. ∫ (e^x+1)^3 e^x dx.

Solution

Here are the detailed solutions to the integrals given in the image:


1. 11xdx\int \frac{1}{1-x} \, dx

Solution: 11xdx=ln1x+C\int \frac{1}{1-x} \, dx = -\ln|1-x| + C


2. 52x+7dx\int \frac{5}{2x+7} \, dx

Solution: Let u=2x+7    du=2dxu = 2x+7 \implies du = 2 \, dx. Rewrite: 52x+7dx=5u12du=52lnu+C\int \frac{5}{2x+7} \, dx = \int \frac{5}{u} \cdot \frac{1}{2} \, du = \frac{5}{2} \ln|u| + C Substitute u=2x+7u = 2x+7: 52x+7dx=52ln2x+7+C\int \frac{5}{2x+7} \, dx = \frac{5}{2} \ln|2x+7| + C


3. 122xx2+4dx\int_{-1}^{2} \frac{2x}{x^2+4} \, dx

Solution: Let u=x2+4    du=2xdxu = x^2+4 \implies du = 2x \, dx. Rewrite: 2xx2+4dx=1udu=lnu+C\int \frac{2x}{x^2+4} \, dx = \int \frac{1}{u} \, du = \ln|u| + C Substitute u=x2+4u = x^2+4: 122xx2+4dx=ln(x2+4)12\int_{-1}^2 \frac{2x}{x^2+4} \, dx = \ln(x^2+4) \bigg|_{-1}^2 Evaluate: ln(22+4)ln((1)2+4)=ln(8)ln(5)=ln(85)\ln(2^2+4) - \ln((-1)^2+4) = \ln(8) - \ln(5) = \ln\left(\frac{8}{5}\right)


4. 0ln2e3xdx\int_{0}^{\ln 2} e^{-3x} \, dx

Solution: e3xdx=e3x3+C=13e3x+C\int e^{-3x} \, dx = \frac{e^{-3x}}{-3} + C = -\frac{1}{3}e^{-3x} + C Evaluate from 00 to ln2\ln 2: 0ln2e3xdx=[13e3x]0ln2\int_{0}^{\ln 2} e^{-3x} \, dx = \left[-\frac{1}{3}e^{-3x}\right]_0^{\ln 2} Substitute: 13e3(ln2)+13e0=13123+13=124+13=724-\frac{1}{3}e^{-3(\ln 2)} + \frac{1}{3}e^{0} = -\frac{1}{3}\frac{1}{2^3} + \frac{1}{3} = -\frac{1}{24} + \frac{1}{3} = \frac{7}{24}


5. e1/xx2dx\int \frac{e^{1/x}}{x^2} \, dx

Solution: Let u=1x    du=1x2dxu = \frac{1}{x} \implies du = -\frac{1}{x^2} \, dx. Rewrite: e1/xx2dx=eudu=eu+C\int \frac{e^{1/x}}{x^2} \, dx = -\int e^u \, du = -e^u + C Substitute u=1xu = \frac{1}{x}: e1/xx2dx=e1/x+C\int \frac{e^{1/x}}{x^2} \, dx = -e^{1/x} + C


6. (ex+1)3exdx\int (e^x+1)^3 e^x \, dx

Solution: Let u=ex+1    du=exdxu = e^x+1 \implies du = e^x \, dx. Rewrite: (ex+1)3exdx=u3du=u44+C\int (e^x+1)^3 e^x \, dx = \int u^3 \, du = \frac{u^4}{4} + C Substitute u=ex+1u = e^x+1: (ex+1)3exdx=(ex+1)44+C\int (e^x+1)^3 e^x \, dx = \frac{(e^x+1)^4}{4} + C


Would you like additional explanations or numerical evaluations?

Here are 5 related questions for practice:

  1. How would you approach integrals involving trigonometric substitutions?
  2. Can you explain how to handle improper integrals with infinite limits?
  3. What are some examples of integrals that require partial fraction decomposition?
  4. How do substitution and integration by parts compare in terms of application?
  5. Can you show an example of solving a definite integral geometrically?

Tip: Always check if simplifying the integrand through substitution makes the integration process smoother.

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Math Problem Analysis

Mathematical Concepts

Integration techniques
Substitution method
Definite and indefinite integrals
Logarithmic functions
Exponential functions

Formulas

∫ 1/(a+bx) dx = (1/b) ln|a+bx| + C
∫ e^(ax) dx = (1/a) e^(ax) + C
∫ x/(x^2 + a^2) dx = (1/2) ln(x^2 + a^2) + C
Definite integral evaluation: ∫(from a to b) f(x) dx = F(b) - F(a)

Theorems

Fundamental Theorem of Calculus
Substitution Rule for Integration

Suitable Grade Level

Undergraduate Level (Calculus I/II)