Problem 15 - Essential Calculus

Question

Find the integral. (Note: Solve by the simplest method-not all require integration by parts.) $$ \int \frac{x e^{2 x}}{(2 x+1)^{2}} d x $$

Step-by-Step Solution

Answer

\(\frac{1}{2}(e^{2x+1} - Ei(2x+1))\)

Step 1: Identify Proper Substitution

One will note that the denominator has the form \(u^2\) with \(u = 2x+1\), thus we will set \(u = 2x + 1\). Then we differentiate \(u\) to get the differential du. We get \(du = 2dx\) or \(dx = \frac{du}{2}\)

Step 2: Substitute into the Integral

Substitute \(u\) and \(dx\) into the integral. This transforms the integral to \(\int \frac{(u-1)e^{u}}{(u)^{2}} \frac{du}{2}\). This can be simplified to \(\frac{1}{2} \int \frac{(u-1)e^{u}}{u^{2}} du\)

Step 3: Split into Separate Integrals

The last expression can be separated into two simpler integrals, rewriting as \(\frac{1}{2} \int \frac{ue^{u}}{u^{2}} du - \frac{1}{2} \int \frac{e^{u}}{u^{2}} du\), which further simplifies to \(\frac{1}{2} \int \frac{e^{u}}{u} du - \frac{1}{2} \int \frac{e^{u}}{u^{2}} du\)

Step 4: Evaluate the Integrals

The first integral is a basic form, \( \int \frac{e^{u}}{u} du\), which evaluates to \(Ei(u)\) where \(Ei(u)\) is the exponential integral. The second integral can be solved by integration by parts. Let \(v = e^{u}\) and \(dw = \frac{1}{u^2}du\). Then \(dv = e^{u}du\) and \(w = -\frac{1}{u}\). Applying the integration by parts formula, we get the second integral to be \(-e^{u}- \int \frac{e^{u}}{u} du\) which simplifies to \(-e^{u} - Ei(u)\)

Step 5: Replace u with Original Variable

Substitute original variable back \(u=2x+1\) into the integral, then we get the final answer as \(\frac{1}{2}(Ei(2x+1) - e^{2x+1} - Ei(2x+1)) = \frac{1}{2}(e^{2x+1} - Ei(2x+1))\).