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- https://math.libretexts.org/Courses/Mission_College/Math_3B%3A_Calculus_II_(Reed)/08%3A_Techniques_of_Integration/8.08%3A_Chapter_8_Review_Exercises\(\displaystyle ∫\frac{3x}{x^3+2x^2−5x−6}\,dx = \frac{1}{10}\big(4\ln|2−x|+5\ln|x+1|−9\ln|x+3|\big)+C\) \(\displaystyle ∫\frac{1}{x^4+4}\,dx = \frac{1}{16}\ln(\frac{x^2+2x+2}{x^2−2x+2})−\frac{1}{8}\ta...\(\displaystyle ∫\frac{3x}{x^3+2x^2−5x−6}\,dx = \frac{1}{10}\big(4\ln|2−x|+5\ln|x+1|−9\ln|x+3|\big)+C\) \(\displaystyle ∫\frac{1}{x^4+4}\,dx = \frac{1}{16}\ln(\frac{x^2+2x+2}{x^2−2x+2})−\frac{1}{8}\tan^{−1}(1−x)+\frac{1}{8}\tan^{−1}(x+1)+C\) 24) [T] Using your function from the previous problem, find exactly how far the Bugati Veyron traveled in the 1 min 40 sec included in the graph.
- https://math.libretexts.org/Courses/Mission_College/Mission_College_MAT_003B/03%3A_Techniques_of_Integration/3.08%3A_Chapter_7_Review_Exercises\(\displaystyle ∫\frac{3x}{x^3+2x^2−5x−6}\,dx = \frac{1}{10}\big(4\ln|2−x|+5\ln|x+1|−9\ln|x+3|\big)+C\) \(\displaystyle ∫\frac{1}{x^4+4}\,dx = \frac{1}{16}\ln(\frac{x^2+2x+2}{x^2−2x+2})−\frac{1}{8}\ta...\(\displaystyle ∫\frac{3x}{x^3+2x^2−5x−6}\,dx = \frac{1}{10}\big(4\ln|2−x|+5\ln|x+1|−9\ln|x+3|\big)+C\) \(\displaystyle ∫\frac{1}{x^4+4}\,dx = \frac{1}{16}\ln(\frac{x^2+2x+2}{x^2−2x+2})−\frac{1}{8}\tan^{−1}(1−x)+\frac{1}{8}\tan^{−1}(x+1)+C\) 24) [T] Using your function from the previous problem, find exactly how far the Bugati Veyron traveled in the 1 min 40 sec included in the graph.
- https://math.libretexts.org/Workbench/MAT_2420_Calculus_II/03%3A_Techniques_of_Integration/3.09%3A_Review_Exercises\(\displaystyle ∫\frac{3x}{x^3+2x^2−5x−6}\,dx = \frac{1}{10}\big(4\ln|2−x|+5\ln|x+1|−9\ln|x+3|\big)+C\) \(\displaystyle ∫\frac{1}{x^4+4}\,dx = \frac{1}{16}\ln(\frac{x^2+2x+2}{x^2−2x+2})−\frac{1}{8}\ta...\(\displaystyle ∫\frac{3x}{x^3+2x^2−5x−6}\,dx = \frac{1}{10}\big(4\ln|2−x|+5\ln|x+1|−9\ln|x+3|\big)+C\) \(\displaystyle ∫\frac{1}{x^4+4}\,dx = \frac{1}{16}\ln(\frac{x^2+2x+2}{x^2−2x+2})−\frac{1}{8}\tan^{−1}(1−x)+\frac{1}{8}\tan^{−1}(x+1)+C\) 24) [T] Using your function from the previous problem, find exactly how far the Bugati Veyron traveled in the 1 min 40 sec included in the graph.
- https://math.libretexts.org/Courses/Mission_College/MAT_3B_Calculus_II_(Kravets)/08%3A_Techniques_of_Integration/8.08%3A_Chapter_8_Review_Exercises\(\displaystyle ∫\frac{3x}{x^3+2x^2−5x−6}\,dx = \frac{1}{10}\big(4\ln|2−x|+5\ln|x+1|−9\ln|x+3|\big)+C\) \(\displaystyle ∫\frac{1}{x^4+4}\,dx = \frac{1}{16}\ln(\frac{x^2+2x+2}{x^2−2x+2})−\frac{1}{8}\ta...\(\displaystyle ∫\frac{3x}{x^3+2x^2−5x−6}\,dx = \frac{1}{10}\big(4\ln|2−x|+5\ln|x+1|−9\ln|x+3|\big)+C\) \(\displaystyle ∫\frac{1}{x^4+4}\,dx = \frac{1}{16}\ln(\frac{x^2+2x+2}{x^2−2x+2})−\frac{1}{8}\tan^{−1}(1−x)+\frac{1}{8}\tan^{−1}(x+1)+C\) 24) [T] Using your function from the previous problem, find exactly how far the Bugati Veyron traveled in the 1 min 40 sec included in the graph.
- https://math.libretexts.org/Courses/Lake_Tahoe_Community_College/Interactive_Calculus_Q2/03%3A_Techniques_of_Integration/3.09%3A_Chapter_7_Review_Exercises\(\displaystyle ∫\frac{3x}{x^3+2x^2−5x−6}\,dx = \frac{1}{10}\big(4\ln|2−x|+5\ln|x+1|−9\ln|x+3|\big)+C\) \(\displaystyle ∫\frac{1}{x^4+4}\,dx = \frac{1}{16}\ln(\frac{x^2+2x+2}{x^2−2x+2})−\frac{1}{8}\ta...\(\displaystyle ∫\frac{3x}{x^3+2x^2−5x−6}\,dx = \frac{1}{10}\big(4\ln|2−x|+5\ln|x+1|−9\ln|x+3|\big)+C\) \(\displaystyle ∫\frac{1}{x^4+4}\,dx = \frac{1}{16}\ln(\frac{x^2+2x+2}{x^2−2x+2})−\frac{1}{8}\tan^{−1}(1−x)+\frac{1}{8}\tan^{−1}(x+1)+C\) 24) [T] Using your function from the previous problem, find exactly how far the Bugati Veyron traveled in the 1 min 40 sec included in the graph.
- https://math.libretexts.org/Courses/Mission_College/Math_3B%3A_Calculus_2_(Sklar)/07%3A_Techniques_of_Integration/7.08%3A_Chapter_7_Review_Exercises\(\displaystyle ∫\frac{3x}{x^3+2x^2−5x−6}\,dx = \frac{1}{10}\big(4\ln|2−x|+5\ln|x+1|−9\ln|x+3|\big)+C\) \(\displaystyle ∫\frac{1}{x^4+4}\,dx = \frac{1}{16}\ln(\frac{x^2+2x+2}{x^2−2x+2})−\frac{1}{8}\ta...\(\displaystyle ∫\frac{3x}{x^3+2x^2−5x−6}\,dx = \frac{1}{10}\big(4\ln|2−x|+5\ln|x+1|−9\ln|x+3|\big)+C\) \(\displaystyle ∫\frac{1}{x^4+4}\,dx = \frac{1}{16}\ln(\frac{x^2+2x+2}{x^2−2x+2})−\frac{1}{8}\tan^{−1}(1−x)+\frac{1}{8}\tan^{−1}(x+1)+C\) 24) [T] Using your function from the previous problem, find exactly how far the Bugati Veyron traveled in the 1 min 40 sec included in the graph.
- https://math.libretexts.org/Courses/SUNY_Geneseo/Math_222_Calculus_2/02%3A_Techniques_of_Integration/2.08%3A_Chapter_2_Review_Exercises\(\displaystyle ∫\frac{3x}{x^3+2x^2−5x−6}\,dx = \frac{1}{10}\big(4\ln|2−x|+5\ln|x+1|−9\ln|x+3|\big)+C\) \(\displaystyle ∫\frac{1}{x^4+4}\,dx = \frac{1}{16}\ln(\frac{x^2+2x+2}{x^2−2x+2})−\frac{1}{8}\ta...\(\displaystyle ∫\frac{3x}{x^3+2x^2−5x−6}\,dx = \frac{1}{10}\big(4\ln|2−x|+5\ln|x+1|−9\ln|x+3|\big)+C\) \(\displaystyle ∫\frac{1}{x^4+4}\,dx = \frac{1}{16}\ln(\frac{x^2+2x+2}{x^2−2x+2})−\frac{1}{8}\tan^{−1}(1−x)+\frac{1}{8}\tan^{−1}(x+1)+C\) 24) [T] Using your function from the previous problem, find exactly how far the Bugati Veyron traveled in the 1 min 40 sec included in the graph.
