Math Problem Statement

Aplicando el método de sustitución, hallar las siguientes integrales indefinidas.

Solution

Let's go through the integrals one by one using substitution as requested:

  1. x4(1+x5)8dx\int x^4(1 + x^5)^8 \, dx

    • Substitute: Let u=1+x5u = 1 + x^5, hence du=5x4dxdu = 5x^4 \, dx.
    • The integral simplifies to: 15u8du=15u99+C=(1+x5)945+C\frac{1}{5} \int u^8 \, du = \frac{1}{5} \cdot \frac{u^9}{9} + C = \frac{(1 + x^5)^9}{45} + C
  2. 2xcos(x2)dx\int 2x \cos(x^2) \, dx

    • Substitute: Let u=x2u = x^2, hence du=2xdxdu = 2x \, dx.
    • The integral becomes: cos(u)du=sin(u)+C=sin(x2)+C\int \cos(u) \, du = \sin(u) + C = \sin(x^2) + C
  3. x2+x2dx\int \frac{x}{\sqrt{2 + x^2}} \, dx

    • Substitute: Let u=2+x2u = 2 + x^2, hence du=2xdxdu = 2x \, dx.
    • The integral simplifies to: 12duu=u+C=2+x2+C\frac{1}{2} \int \frac{du}{\sqrt{u}} = \sqrt{u} + C = \sqrt{2 + x^2} + C
  4. ln(x)xdx\int \frac{\ln(x)}{x} \, dx

    • Substitute: Let u=ln(x)u = \ln(x), hence du=1xdxdu = \frac{1}{x} \, dx.
    • The integral becomes: udu=u22+C=ln2(x)2+C\int u \, du = \frac{u^2}{2} + C = \frac{\ln^2(x)}{2} + C
  5. sin(x)xdx\int \frac{\sin(\sqrt{x})}{\sqrt{x}} \, dx

    • Substitute: Let u=xu = \sqrt{x}, hence du=12xdxdu = \frac{1}{2\sqrt{x}} \, dx.
    • The integral becomes: 2sin(u)du=2cos(u)+C=2cos(x)+C2 \int \sin(u) \, du = -2 \cos(u) + C = -2 \cos(\sqrt{x}) + C
  6. xcos(4x2)dx\int x \cos(4x^2) \, dx

    • Substitute: Let u=4x2u = 4x^2, hence du=8xdxdu = 8x \, dx.
    • The integral becomes: 18cos(u)du=18sin(u)+C=sin(4x2)8+C\frac{1}{8} \int \cos(u) \, du = \frac{1}{8} \sin(u) + C = \frac{\sin(4x^2)}{8} + C
  7. 1xln(x)dx\int \frac{1}{x \ln(x)} \, dx

    • Substitute: Let u=ln(x)u = \ln(x), hence du=1xdxdu = \frac{1}{x} \, dx.
    • The integral becomes: 1udu=lnu+C=lnln(x)+C\int \frac{1}{u} \, du = \ln|u| + C = \ln|\ln(x)| + C
  8. tan(x)dx\int \tan(x) \, dx

    • Recall that tan(x)dx=lnsec(x)+C\int \tan(x) \, dx = \ln|\sec(x)| + C.
  9. x2ln(x)dx\int x^2 \ln(x) \, dx

    • Integration by parts: Let u=ln(x)u = \ln(x), dv=x2dxdv = x^2 \, dx, hence du=1xdxdu = \frac{1}{x} \, dx, v=x33v = \frac{x^3}{3}.
    • The integral becomes: x3ln(x)3x331xdx=x3ln(x)3x39+C\frac{x^3 \ln(x)}{3} - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx = \frac{x^3 \ln(x)}{3} - \frac{x^3}{9} + C
  10. exxdx\int \frac{e^{\sqrt{x}}}{\sqrt{x}} \, dx

  • Substitute: Let u=xu = \sqrt{x}, hence du=12xdxdu = \frac{1}{2\sqrt{x}} \, dx.
  • The integral becomes: 2eudu=2eu+C=2ex+C2 \int e^u \, du = 2e^u + C = 2e^{\sqrt{x}} + C

Let me know if you'd like further explanations or if anything is unclear!

Here are 5 related questions:

  1. How is substitution chosen in integration problems?
  2. How do you recognize when to use integration by parts?
  3. What is the difference between indefinite and definite integrals?
  4. How do you integrate functions involving logarithms?
  5. Can you apply the substitution method to all integrals?

Tip: Always verify your substitution by differentiating your result to check if you recover the original integrand.

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Math Problem Analysis

Mathematical Concepts

Integral Calculus
Substitution Method
Integration by Parts
Trigonometric Functions
Logarithmic Functions
Exponential Functions

Formulas

u-substitution: u = f(x), du = f'(x) dx
Integration by parts: ∫u dv = uv - ∫v du
Basic integrals: ∫x^n dx = (x^(n+1))/(n+1) + C

Theorems

Fundamental Theorem of Calculus
Substitution Rule for Integration

Suitable Grade Level

Undergraduate (Calculus)