Michael Bowen's VC Course Pages

Answers

Math V21B Section 7.5

Introduction

Please try this problem on your own before looking at my solution. There may be other ways to solve this; I make no claim that this method is the quickest, prettiest, or most efficient.

Problem 49

The original problem is $$I = \int {\frac{1}{{x\sqrt {4x + 1} }}\;dx} $$

From the square root, we suspect a trigonometric substitution might be the best method, but there is no squared variable inside. However, we can make this happen with a clever $u$-substitution: $$u = 2\sqrt x = 2{x^{{1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-1.2pt} 2}}} \to \frac{u}{2} = \frac{{2\sqrt x }}{2} = \sqrt x $$ $${u^2} = {\left( {2\sqrt x } \right)^2} = 4x \to {u^2} + 1 = 4x + 1 \to \sqrt {{u^2} + 1} = \sqrt {4x + 1} $$ $$du = 2\left( {\frac{1}{2}{x^{ - {1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-1.2pt} 2}}}} \right)\;dx = {x^{ - {1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-1.2pt} 2}}}\;dx = \frac{1}{{\sqrt x }}\;dx$$ Rewriting the original problem makes it more clear how to place the above substitutions: $$I = \int {\frac{1}{{x\sqrt {4x + 1} }}\;dx} = \int {\frac{1}{{\sqrt x \sqrt x \sqrt {4x + 1} }}\;dx} = \int {\frac{1}{{\sqrt x }} \cdot \frac{1}{{\sqrt {4x + 1} }} \cdot \frac{1}{{\sqrt x }}\;dx} $$ $$I = \int {\frac{1}{\left({\tfrac{u}{2}}\right)} \cdot \frac{1}{{\sqrt {{u^2} + 1} }} \cdot du} = \int {\frac{2}{u} \cdot \frac{1}{{\sqrt {{u^2} + 1} }} \cdot du} = 2\int {\frac{1}{u} \cdot \frac{1}{{\sqrt {{u^2} + 1} }} \cdot du} $$

Now we can apply the trigonometric substitution; with $$u = \tan \;\theta ;\quad du = {\sec ^2}\theta \;d\theta $$ we can write $$I = 2\int {\frac{1}{{\tan \;\theta }} \cdot \frac{1}{{\sqrt {{{\tan }^2}\theta + 1} }} \cdot {{\sec }^2}\theta \;d\theta } = 2\int {\frac{1}{{\tan \;\theta }} \cdot \frac{1}{{\sqrt {{{\sec }^2}\theta } }} \cdot {{\sec }^2}\theta \;d\theta } = 2\int {\frac{1}{{\tan \;\theta }}\frac{{{{\sec }^2}\theta }}{{\sec \;\theta }}\;d\theta } = 2\int {\frac{{\sec \;\theta }}{{\tan \;\theta }}\;d\theta } $$ $$I = 2\int {\frac{{1/\cos \;\theta }}{{\sin \;\theta /\cos \;\theta }}\;d\theta } = 2\int {\frac{1}{{\sin \;\theta }}\;d\theta } = 2\int {\csc \;\theta \;d\theta } $$

The solution to this last integral appears in the brief table on page 484 of the textbook: $$I = 2\,\ln \big\lvert {\csc \;\theta - \cot \;\theta } \big\rvert + C$$ From here, it is necessary to back-substitute; first for $u$ (using a right triangle), then for $x$: $$I = 2\,\ln \Bigg\lvert {\frac{{\sqrt {{u^2} + 1} }}{u} - \frac{1}{u}} \Bigg\rvert + C = 2\,\ln \Bigg\lvert {\frac{{\sqrt {{{\left( {2\sqrt x } \right)}^2} + 1} }}{{2\sqrt x }} - \frac{1}{{2\sqrt x }}} \Bigg\rvert + C = \underline{\underline {2\,\ln \Bigg\lvert {\frac{{\sqrt {4x + 1} - 1}}{{2\sqrt x }}} \Bigg\rvert + C}} $$ The underlined answer at the end of the last line would suffice on an exam; however, it differs from the textbook's answer. As usual, additional algebra is required to recover the textbook's answer (which is equivalent to this one).

One way to get there is to use logarithm properties to bring the coefficient 2 inside the logarithm's argument as a power: $$I = \ln {\Bigg\lvert {\frac{{\sqrt {4x + 1} - 1}}{{2\sqrt x }}} \Bigg\rvert^2} + C = \ln \Bigg\lvert {\frac{{\left( {\sqrt {4x + 1} - 1} \right)\left( {\sqrt {4x + 1} - 1} \right)}}{{4x}}} \Bigg\rvert + C$$ We can then rationalize one of the matching factors in the numerator by multiplying the fraction, top and bottom, with the conjugate of this factor: $$I = \ln \Bigg\lvert {\frac{{\left( {\sqrt {4x + 1} - 1} \right)\left( {\sqrt {4x + 1} - 1} \right)\left( {\sqrt {4x + 1} + 1} \right)}}{{\left( 4x \right)\left( {\sqrt {4x + 1} + 1} \right)}}} \Bigg\rvert + C$$ Distributing only the last two parenthesized factors in the numerator yields $$ I = \ln \Bigg\lvert {\frac{{\left( {\sqrt {4x + 1} - 1} \right)\left[ {\left( {4x + 1} \right) - 1} \right]}}{{\left( 4x \right)\left( {\sqrt {4x + 1} + 1} \right)}}} \Bigg\rvert + C = \ln \Bigg\lvert {\frac{{\left( {\sqrt {4x + 1} - 1} \right)\left[ \cancel{4x} \right]}}{{\left( \cancel{4x} \right)\left( {\sqrt {4x + 1} + 1} \right)}}} \Bigg\rvert + C $$ $$ \boxed{\displaystyle{I = \ln \Bigg\lvert {\frac{{\left( {\sqrt {4x + 1} - 1} \right)}}{{\left( {\sqrt {4x + 1} + 1} \right)}}} \Bigg\rvert + C}} $$ which agrees in all essential regards with the solution in the back of the textbook.