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Chapter 1: Problem 1

Evaluate the following integrals: $$ \int \frac{x d x}{x^{2}+2 x+1} $$

Short Answer

Expert verified
Question: Evaluate the integral $\int \frac{x dx}{x^{2}+2 x+1}$. Answer: $\ln(|x+1|) + \frac{1}{x+1} + C$

Step by step solution

01

Analyze the integrand

First, we analyze the integrand, a rational function. In particular, we want to simplify the denominator: $$ x^2+2x+1 $$ Notice that it is a perfect square of \((x+1)^2\). Therefore, the integrand can be simplified to: $$ \frac{x}{(x+1)^2} $$
02

Use substitution

To solve the integral, we can use substitution. Let's define a new variable, \(u\), as: $$ u=x+1 $$ Now, we take the derivative of \(u\) with respect to \(x\) and obtain \(d u = d x\).
03

Rewrite the integral in terms of u

Now, we can rewrite the integral in terms of the new variable \(u\): $$ \int \frac{x dx}{(x+1)^2} = \int\frac{(u-1) du}{u^2} $$
04

Divide the fraction

To continue, we will divide the fraction \((u-1)/u^2\), so we separate the terms in the numerator: $$ \int \frac{(u-1)du}{u^2} = \int \left(\frac{u}{u^2} - \frac{1}{u^2}\right)du $$ We then simplify before integrating: $$ \int \left(\frac{1}{u} - \frac{1}{u^2}\right)du $$
05

Integrate

Now, we integrate each term of the integrand separately: $$ \int \left(\frac{1}{u} - \frac{1}{u^2}\right)du = \int \frac{1}{u} du - \int\frac{1}{u^2}du $$ Using basic integration rules, we find: $$ \ln(|u|) + C_1 - \left(-\frac{1}{u} + C_2\right) $$ Combining the constants, we have: $$ \ln(|u|) + \frac{1}{u} + C $$
06

Substitute back for x

Finally, we substitute back the original variable, \(x\), using our previous substitution \(u = x + 1\): $$ \ln(|x+1|) + \frac{1}{x+1} + C $$ So, the solution to the integral is: $$ \int \frac{x dx}{x^{2}+2 x+1}=\ln(|x+1|) + \frac{1}{x+1} + C $$

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Most popular questions from this chapter

Evaluate the following integrals: (i) \(\int \frac{\sqrt{x^{4}+x^{-4}+2}}{x^{3}} d x\) (ii) \(\int \frac{d x}{\sqrt{2 x+3}+\sqrt{2 x-3}} d x\) (iii) \(\int \frac{(\sqrt{x}+1)\left(x^{2}-\sqrt{x}\right)}{x \sqrt{x}+x+\sqrt{x}} d x\) (iv) \(\int\left(\frac{1-x^{-2}}{x^{1 / 2}-x^{-1 / 2}}-\frac{2}{x^{3 / 2}}+\frac{x^{-2}-x}{x^{1 / 2}-x^{-1 / 2}}\right) d x\)

\(\int \frac{\sqrt{x^{2}+1}}{x^{4}} \ln \left(1+\frac{1}{x^{2}}\right) d x\)

Evaluate the following integrals:(i) \(\int \frac{1}{(\cos x+2 \sin x)^{2}} d x\) (ii) \(\int \frac{\mathrm{dx}}{\left(\sin ^{2} \mathrm{x}+2 \cos ^{2} \mathrm{x}\right)^{2}} \mathrm{dx}\) (iii) \(\int \frac{\cos \theta \mathrm{d} \theta}{(5+4 \cos \theta)^{2}}\) (iv) \(\int \frac{d x}{\sin ^{6} x+\cos ^{6} x}\)

Evaluate the following integrals : (i) \(\int \mathrm{e}^{\mathrm{x}}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}^{2}}\right)^{2} \mathrm{dx}\) (ii) \(\int \mathrm{e}^{x} \frac{\left(x^{3}-x+2\right)}{\left(x^{2}+1\right)^{2}} d x\) (iii) \(\int \frac{\mathrm{e}^{\mathrm{x}}(\mathrm{x}-1)}{(\mathrm{x}+1)^{3}} \mathrm{dx}\) (iv) \(\int \mathrm{e}^{x}\left(\frac{1-x}{1+x}\right)^{2} d x\)

Only one of the functions \(\sqrt{x} \sqrt[3]{1-x}\) and \(\sqrt{1-x} \sqrt[3]{1-x}\) has an elementary antiderivative. Find the function.
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