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Chapter 6: Problem 15

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

Short Answer

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

Step by step solution

01

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 \(dx = \frac{du}{2}\)
02

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 \(\frac{1}{2} \int \frac{(u-1)e^{u}}{u^{2}} du\)
03

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 \(\frac{1}{2} \int \frac{e^{u}}{u} du - \frac{1}{2} \int \frac{e^{u}}{u^{2}} du\)
04

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 \(-e^{u} - Ei(u)\)
05

Replace u with Original Variable

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

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Most popular questions from this chapter

Find the integral. Use a computer algebra system to confirm your result. $$ \int \csc ^{2} 3 x \cot 3 x d x $$

Given continuous functions \(f\) and \(g\) such that \(0 \leq f(x) \leq g(x)\) on the interval \([a, \infty),\) prove the following. (a) If \(\int_{a}^{\infty} g(x) d x\) converges, then \(\int_{a}^{\infty} f(x) d x\) converges. (b) If \(\int_{a}^{\infty} f(x) d x\) diverges, then \(\int_{a}^{\infty} g(x) d x\) diverges.

Sketch the graph of \(g(x)=\left\\{\begin{array}{ll}e^{-1 / x^{2}}, & x \neq 0 \\ 0, & x=0\end{array}\right.\) and determine \(g^{\prime}(0)\).

Prove the following generalization of the Mean Value Theorem. If \(f\) is twice differentiable on the closed interval \([a, b],\) then \(f(b)-f(a)=f^{\prime}(a)(b-a)-\int_{a}^{b} f^{\prime \prime}(t)(t-b) d t\).

For what value of \(c\) does the integral \(\int_{0}^{\infty}\left(\frac{1}{\sqrt{x^{2}+1}}-\frac{c}{x+1}\right) d x\) converge? Evaluate the integral for this value of \(c\).
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