Math Problem Statement

e^x(1+x)(x+lnx)dx

Solution

To solve the integral:

ex(1+x)(x+lnx)dx\int e^x(1 + x)(x + \ln x) \, dx

we need to approach it step by step.

Step 1: Expand the expression inside the integral.

We begin by expanding the product (1+x)(x+lnx)(1 + x)(x + \ln x):

(1+x)(x+lnx)=1(x+lnx)+x(x+lnx)(1 + x)(x + \ln x) = 1 \cdot (x + \ln x) + x \cdot (x + \ln x) =x+lnx+x2+xlnx= x + \ln x + x^2 + x \ln x

Now the integral becomes:

ex(x+lnx+x2+xlnx)dx\int e^x \left( x + \ln x + x^2 + x \ln x \right) \, dx

Step 2: Break the integral into separate parts.

We can now separate the integral into simpler parts:

exxdx+exlnxdx+exx2dx+exxlnxdx\int e^x x \, dx + \int e^x \ln x \, dx + \int e^x x^2 \, dx + \int e^x x \ln x \, dx

Step 3: Solve the integrals.

Each of these integrals involves the exponential function exe^x, which suggests that integration by parts will be helpful. Let's solve these one by one.

1. exxdx\int e^x x \, dx

Using integration by parts, where we set u=xu = x and dv=exdxdv = e^x \, dx, we have:

du=dx,v=exdu = dx, \quad v = e^x exxdx=xexexdx=xexex+C1\int e^x x \, dx = x e^x - \int e^x \, dx = x e^x - e^x + C_1

2. exlnxdx\int e^x \ln x \, dx

This one requires integration by parts, where we set u=lnxu = \ln x and dv=exdxdv = e^x \, dx. Then:

du=1xdx,v=exdu = \frac{1}{x} dx, \quad v = e^x exlnxdx=lnxexexxdx\int e^x \ln x \, dx = \ln x \cdot e^x - \int \frac{e^x}{x} \, dx

This integral doesn't have a simple elementary form (the second integral exxdx\int \frac{e^x}{x} \, dx is the Exponential Integral, denoted by Ei(x)\text{Ei}(x)).

3. exx2dx\int e^x x^2 \, dx

Using integration by parts again, where we set u=x2u = x^2 and dv=exdxdv = e^x \, dx, we get:

du=2xdx,v=exdu = 2x \, dx, \quad v = e^x exx2dx=x2ex2xexdx\int e^x x^2 \, dx = x^2 e^x - \int 2x e^x \, dx

We already know how to solve exxdx\int e^x x \, dx, so substitute that result here:

=x2ex2(xexex)= x^2 e^x - 2(x e^x - e^x) =ex(x22x+2)+C2= e^x (x^2 - 2x + 2) + C_2

4. exxlnxdx\int e^x x \ln x \, dx

Again using integration by parts with u=xlnxu = x \ln x and dv=exdxdv = e^x \, dx, we have:

du=lnxdx+1xdx,v=exdu = \ln x \, dx + \frac{1}{x} dx, \quad v = e^x exxlnxdx=xlnxexex(lnx+1x)dx\int e^x x \ln x \, dx = x \ln x \cdot e^x - \int e^x \left( \ln x + \frac{1}{x} \right) \, dx

This is another complex integral involving the Exponential Integral function.

Step 4: Combine the results.

Combining all the results, the integral becomes:

ex(1+x)(x+lnx)dx=(Result from the expansion above)\int e^x(1 + x)(x + \ln x) \, dx = \text{(Result from the expansion above)}

However, parts of the integral (involving lnx\ln x and exx\frac{e^x}{x}) lead to special functions that are outside elementary functions.

Would you like a deeper breakdown or further clarification on any of the steps above? Or perhaps an estimate for the remaining terms?

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Math Problem Analysis

Mathematical Concepts

Integral Calculus
Integration by Parts
Exponential Functions
Special Functions

Formulas

Integration by Parts: ∫udv = uv - ∫vdu
Exponential Integral: Ei(x)

Theorems

Integration by Parts Theorem

Suitable Grade Level

Grades 11-12 (Advanced)