Evaluate: \int \frac{tan^{-1} 9x}{ 1 +81 x^2} dx \\ \int \frac{x^9}{1 + x^{20}} dx\\ \int_1^2...

Question:

Evaluate:

{eq}\int \frac{tan^{-1} 9x}{ 1 +81 x^2} dx \\ \int \frac{x^9}{1 + x^{20}} dx\\ \int_1^2 \frac{e^{1/x^5}}{x^6}dx \\ \int_9^{10} x \sqrt{x-9} dx{/eq}

Integration


We will use the substitution method to solve the integrals presented in the questions. We will add a constant C after performing the required integration.

The required steps to evaluate{eq}\int f(x)\quad dx {/eq}by substitution are:

1. Put x=g(y), dx = g'(y) dy in the integrand

2. Evaluate the resulting integral in y.

3. Express the result obtained in terms of x.

Answer and Explanation:

1) {eq}\displaystyle{\int \frac{tan^{-1} 9x}{ 1 +81 x^2} dx \\[10pt] \text{substitute}\quad u = tan^{-1} 9x \\[10pt] \frac{du}{dx} = \frac{1}{(9x)^2 +1}\frac{d}{dx}(9x)\hspace{100pt}\left(\frac{d}{dx} tan^{-1} x = \frac{1}{1 +x^2}\right)\\[10pt] \frac{du}{dx} = \frac{9}{1 + 81x^2}\\[10pt] dx = \frac{81x^2 +1}{9} \ du\\[10pt] \int \frac{u}{9} \ du\\[10pt] \frac{1}{9}\int u \ du\\[10pt] \left(\frac{u^2}{2}\right) + C\\[10pt] \frac{u^2}{18} + C\\[10pt] \boxed{\frac{(tan^{-1} 9x)^2}{18} + C}\\} {/eq}


2) {eq}\displaystyle{\int \frac{x^9}{1 + x^{20}} dx\\[10pt] \text{substitute}\quad u = x^{10} \\[10pt] \frac{du}{dx} = 10 x^9\\[10pt] dx = \frac{1}{10x^9} \ du\\[10pt] \frac{1}{10}\int \frac{1}{1 +u^2} \ du\\[10pt] \frac{1}{10} \tan^{-1} u + C\\[10pt] \boxed{\frac{tan^{-1}(x^{10})}{10} +C}\\} {/eq}


3) {eq}\displaystyle{\int_1^2 \frac{e^{1/x^5}}{x^6}dx \\[10pt] \text{substitute}\quad u = \frac{1}{x^5}\\[10pt] \frac{du}{dx} = \frac{-5x^4}{x^{10}} = \frac{-5}{x^6}\\[10pt] dx = \frac{-x^6}{5}\\[10pt] \frac{-1}{5}\int_1^2 e^u \ du\\[10pt] \frac{-1}{5}\left[ e^u\right]_1^2\\[10pt] \frac{-1}{5}\left[ e^\frac{1}{x^5}\right]_1^2\\[10pt] \left[\frac{-e^\frac{1}{x^5}}{5}\right]_1^2\\[10pt] \left[\frac{-e^\frac{1}{32}}{5}\right] - \left[\frac{-e^{1}}{5}\right]\\[10pt] \boxed{\frac{-\sqrt[32]e + e}{5}}\\} {/eq}


4) {eq}\displaystyle{\int_9^{10} x \sqrt{x-9} dx\\[10pt] \text{substitute}\quad u = x -9 \\[10pt] \frac{du}{dx} = 1\\[10pt] dx = du\\[10pt] \text{Also,}\quad x = u + 9\\[10pt] \int_{9}^{10} (u + 9)\sqrt{u} \ du\\[10pt] \int_{9}^{10} u^\frac{3}{2} + 9 u^\frac{1}{2} \ du\\[10pt] \left[ \frac{2u^\frac{5}{2}}{5} + \frac{18u^\frac{3}{2}}{3}\right]_{9}^{10}\\[10pt] \left[ \frac{2(x - 9)^\frac{5}{2}}{5} + 6(x - 9)^\frac{3}{2}\right]_{9}^{10}\\[10pt] \frac{2(10 - 9)^\frac{5}{2}}{5} + 6(10 - 9)^\frac{3}{2} - 0\\[10pt] \frac{2}{5} + 6\\[10pt] \frac{2 + 30}{5}\\[10pt] \boxed{\frac{32}{5}}\\} {/eq}


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How to Solve Integrals Using Substitution

from Math 104: Calculus

Chapter 13 / Lesson 5
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