Math Problem Statement

Find the indefinite integral.

∫(e^(3x)−x)^2dx

Answer:

Solution

To solve the indefinite integral

(e3xx)2dx,\int (e^{3x} - x)^2 \, dx,

we begin by expanding the integrand.

Step 1: Expand the expression (e3xx)2(e^{3x} - x)^2

(e3xx)2=(e3x)22e3xx+x2=e6x2xe3x+x2.(e^{3x} - x)^2 = (e^{3x})^2 - 2e^{3x} \cdot x + x^2 = e^{6x} - 2x e^{3x} + x^2.

Step 2: Break down the integral into simpler parts

Now we can rewrite the integral as the sum of three integrals:

(e6x2xe3x+x2)dx=e6xdx2xe3xdx+x2dx.\int (e^{6x} - 2x e^{3x} + x^2) \, dx = \int e^{6x} \, dx - 2 \int x e^{3x} \, dx + \int x^2 \, dx.

Step 3: Solve each integral

  1. Integral of e6xe^{6x}: e6xdx=e6x6.\int e^{6x} \, dx = \frac{e^{6x}}{6}.

  2. Integral of xe3xx e^{3x}: To solve xe3xdx\int x e^{3x} \, dx, we use integration by parts. Let: u=xanddv=e3xdx.u = x \quad \text{and} \quad dv = e^{3x} \, dx. Then, du=dxdu = dx and v=e3x3v = \frac{e^{3x}}{3}. Now, apply the integration by parts formula udv=uvvdu\int u \, dv = uv - \int v \, du:

    xe3xdx=xe3x3e3x3dx.\int x e^{3x} \, dx = x \cdot \frac{e^{3x}}{3} - \int \frac{e^{3x}}{3} \, dx.

    The remaining integral is:

    e3x3dx=e3x9.\int \frac{e^{3x}}{3} \, dx = \frac{e^{3x}}{9}.

    Thus:

    xe3xdx=xe3x3e3x9.\int x e^{3x} \, dx = \frac{x e^{3x}}{3} - \frac{e^{3x}}{9}.

  3. Integral of x2x^2: x2dx=x33.\int x^2 \, dx = \frac{x^3}{3}.

Step 4: Combine all results

Now, combine the results of all three integrals:

(e3xx)2dx=e6x62(xe3x3e3x9)+x33.\int (e^{3x} - x)^2 \, dx = \frac{e^{6x}}{6} - 2 \left( \frac{x e^{3x}}{3} - \frac{e^{3x}}{9} \right) + \frac{x^3}{3}.

Simplify the second term:

2(xe3x3e3x9)=2xe3x3+2e3x9.-2 \left( \frac{x e^{3x}}{3} - \frac{e^{3x}}{9} \right) = -\frac{2x e^{3x}}{3} + \frac{2e^{3x}}{9}.

Thus, the final result is:

(e3xx)2dx=e6x62xe3x3+2e3x9+x33+C,\int (e^{3x} - x)^2 \, dx = \frac{e^{6x}}{6} - \frac{2x e^{3x}}{3} + \frac{2e^{3x}}{9} + \frac{x^3}{3} + C,

where CC is the constant of integration.


Would you like further clarification on any step or method? Here are a few related questions:

  1. What is integration by parts, and how does it work?
  2. How do we handle integrals of products like xe3xx e^{3x}?
  3. What is the general approach to expanding expressions before integrating?
  4. How do you determine when to use substitution versus integration by parts?
  5. Can the result be simplified further or expressed in a more compact form?

Tip: When expanding complex terms before integration, always check if any of the parts can be simplified or factored to make the integration process easier.

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Math Problem Analysis

Mathematical Concepts

Integration
Expansion
Integration by Parts
Exponential Functions

Formulas

∫e^(ax)dx = e^(ax)/a
Integration by parts: ∫udv = uv - ∫vdu

Theorems

Integration by parts

Suitable Grade Level

Grades 11-12