Answer on Question #48982 – Math – Integral Calculus
1. Complete the perfect square and use tables of integration to integrate
∫ 1 x 2 − 3 x + 6 d x \int \frac {1}{\sqrt {x ^ {2} - 3 x + 6}} d x ∫ x 2 − 3 x + 6 1 d x
Solution:
∫ 1 x 2 − 3 x + 6 d x = ∫ 1 x 2 − 3 x + 2.25 + 3.75 d x = ∫ 1 ( x − 1.5 ) 2 + 3.75 d x \int \frac {1}{\sqrt {x ^ {2} - 3 x + 6}} d x = \int \frac {1}{\sqrt {x ^ {2} - 3 x + 2 . 2 5 + 3 . 7 5}} d x = \int \frac {1}{\sqrt {(x - 1 . 5) ^ {2} + 3 . 7 5}} d x ∫ x 2 − 3 x + 6 1 d x = ∫ x 2 − 3 x + 2.25 + 3.75 1 d x = ∫ ( x − 1.5 ) 2 + 3.75 1 d x
Use tables of integration:
∫ 1 x 2 + B d x = ln ∣ x + x 2 + B ∣ + C \int \frac {1}{\sqrt {x ^ {2} + B}} d x = \ln \left| x + \sqrt {x ^ {2} + B} \right| + C ∫ x 2 + B 1 d x = ln ∣ ∣ x + x 2 + B ∣ ∣ + C ∫ 1 ( x − 1.5 ) 2 + 3.75 d x = ln ∣ ( x − 1.5 ) + x 2 − 3 x + 6 ∣ + C , C \int \frac{1}{\sqrt{(x - 1.5)^2 + 3.75}} dx = \ln \left| (x - 1.5) + \sqrt{x^2 - 3x + 6} \right| + C, C ∫ ( x − 1.5 ) 2 + 3.75 1 d x = ln ∣ ∣ ( x − 1.5 ) + x 2 − 3 x + 6 ∣ ∣ + C , C
Answer:
∫ 1 x 2 − 3 x + 6 d x = ln ∣ ( x − 1.5 ) + x 2 − 3 x + 6 ∣ + C \int \frac {1}{\sqrt {x ^ {2} - 3 x + 6}} d x = \ln \left| (x - 1. 5) + \sqrt {x ^ {2} - 3 x + 6} \right| + C ∫ x 2 − 3 x + 6 1 d x = ln ∣ ∣ ( x − 1.5 ) + x 2 − 3 x + 6 ∣ ∣ + C
2. Use a substitution technique and then the table of integration to integrate
∫ x 5 x 4 − 4 d x \int x ^ {5} \sqrt {x ^ {4} - 4} d x ∫ x 5 x 4 − 4 d x
Solution:
∫ x 5 x 4 − 4 d x = ∫ x 7 1 − 4 x 4 d x \int x ^ {5} \sqrt {x ^ {4} - 4} d x = \int x ^ {7} \sqrt {1 - \frac {4}{x ^ {4}}} d x ∫ x 5 x 4 − 4 d x = ∫ x 7 1 − x 4 4 d x
Let
1 − 4 x 4 = t 2 , t = x 4 − 4 x 2 1 - \frac {4}{x ^ {4}} = t ^ {2}, t = \frac {\sqrt {x ^ {4} - 4}}{x ^ {2}} 1 − x 4 4 = t 2 , t = x 2 x 4 − 4 x 4 = 4 1 − t 2 x ^ {4} = \frac {4}{1 - t ^ {2}} x 4 = 1 − t 2 4 4 x 3 d x = 8 t ( 1 − t 2 ) 2 d t 4 x ^ {3} d x = \frac {8 t}{(1 - t ^ {2}) ^ {2}} d t 4 x 3 d x = ( 1 − t 2 ) 2 8 t d t d x = 2 t x 3 ( 1 − t 2 ) 2 d t d x = \frac {2 t}{x ^ {3} (1 - t ^ {2}) ^ {2}} d t d x = x 3 ( 1 − t 2 ) 2 2 t d t ∫ x 7 1 − 4 x 4 d x = ∫ x 7 t 2 t x 3 ( 1 − t 2 ) 2 d t = 2 ∫ x 4 t 2 ( 1 − t 2 ) 2 d t = 8 ∫ t 2 ( 1 − t 2 ) 3 d t \int x ^ {7} \sqrt {1 - \frac {4}{x ^ {4}}} d x = \int x ^ {7} t \frac {2 t}{x ^ {3} (1 - t ^ {2}) ^ {2}} d t = 2 \int x ^ {4} \frac {t ^ {2}}{(1 - t ^ {2}) ^ {2}} d t = 8 \int \frac {t ^ {2}}{(1 - t ^ {2}) ^ {3}} d t ∫ x 7 1 − x 4 4 d x = ∫ x 7 t x 3 ( 1 − t 2 ) 2 2 t d t = 2 ∫ x 4 ( 1 − t 2 ) 2 t 2 d t = 8 ∫ ( 1 − t 2 ) 3 t 2 d t
Use the partial fraction:
t 2 ( 1 − t 2 ) 3 = − t 2 ( t 2 − 1 ) 3 = A t + 1 + B ( t + 1 ) 2 + C ( t + 1 ) 3 + D t − 1 + E ( t − 1 ) 2 + F ( t − 1 ) 3 \frac {t ^ {2}}{(1 - t ^ {2}) ^ {3}} = \frac {- t ^ {2}}{(t ^ {2} - 1) ^ {3}} = \frac {A}{t + 1} + \frac {B}{(t + 1) ^ {2}} + \frac {C}{(t + 1) ^ {3}} + \frac {D}{t - 1} + \frac {E}{(t - 1) ^ {2}} + \frac {F}{(t - 1) ^ {3}} ( 1 − t 2 ) 3 t 2 = ( t 2 − 1 ) 3 − t 2 = t + 1 A + ( t + 1 ) 2 B + ( t + 1 ) 3 C + t − 1 D + ( t − 1 ) 2 E + ( t − 1 ) 3 F A ( t + 1 ) 2 ( t − 1 ) 3 + B ( t + 1 ) ( t − 1 ) 3 + C ( t − 1 ) 3 + D ( t − 1 ) 2 ( t + 1 ) 3 + + E ( t − 1 ) ( t + 1 ) 3 + F ( t + 1 ) 3 = − t 2 \begin{array}{l} A (t + 1) ^ {2} (t - 1) ^ {3} + B (t + 1) (t - 1) ^ {3} + C (t - 1) ^ {3} + D (t - 1) ^ {2} (t + 1) ^ {3} + \\ + E (t - 1) (t + 1) ^ {3} + F (t + 1) ^ {3} = - t ^ {2} \\ \end{array} A ( t + 1 ) 2 ( t − 1 ) 3 + B ( t + 1 ) ( t − 1 ) 3 + C ( t − 1 ) 3 + D ( t − 1 ) 2 ( t + 1 ) 3 + + E ( t − 1 ) ( t + 1 ) 3 + F ( t + 1 ) 3 = − t 2 A = − 1 16 , B = − 1 16 , C = 1 8 , D = 1 16 , E = − 1 16 , F = − 1 8 A = - \frac {1}{1 6}, B = - \frac {1}{1 6}, C = \frac {1}{8}, D = \frac {1}{1 6}, E = - \frac {1}{1 6}, F = - \frac {1}{8} A = − 16 1 , B = − 16 1 , C = 8 1 , D = 16 1 , E = − 16 1 , F = − 8 1 8 ∫ t 2 ( 1 − t 2 ) 3 d t = 1 2 ∫ ( − 1 t + 1 − 1 ( t + 1 ) 2 + 2 ( t + 1 ) 3 + 1 t − 1 − 1 ( t − 1 ) 2 − 2 ( t − 1 ) 3 ) d t = = 1 2 ( ( ln t − 1 t + 1 ) + 1 t + 1 + 1 t − 1 − 1 ( t + 1 ) 2 + 1 ( t − 1 ) 2 ) + C = = 1 2 ( ln t − 1 t + 1 ) + t ( t 2 − 1 ) − 2 t ( t 2 − 1 ) 2 + C \begin{array}{l} 8 \int {\frac {t ^ {2}}{(1 - t ^ {2}) ^ {3}}} d t = \frac {1}{2} \int \left(- \frac {1}{t + 1} - \frac {1}{(t + 1) ^ {2}} + \frac {2}{(t + 1) ^ {3}} + \frac {1}{t - 1} - \frac {1}{(t - 1) ^ {2}} - \frac {2}{(t - 1) ^ {3}}\right) d t = \\ = \frac {1}{2} \left(\left(\ln \frac {t - 1}{t + 1}\right) + \frac {1}{t + 1} + \frac {1}{t - 1} - \frac {1}{(t + 1) ^ {2}} + \frac {1}{(t - 1) ^ {2}}\right) + C = \\ = \frac {1}{2} \left(\ln \frac {t - 1}{t + 1}\right) + \frac {t}{\left(t ^ {2} - 1\right)} - \frac {2 t}{\left(t ^ {2} - 1\right) ^ {2}} + C \\ \end{array} 8 ∫ ( 1 − t 2 ) 3 t 2 d t = 2 1 ∫ ( − t + 1 1 − ( t + 1 ) 2 1 + ( t + 1 ) 3 2 + t − 1 1 − ( t − 1 ) 2 1 − ( t − 1 ) 3 2 ) d t = = 2 1 ( ( ln t + 1 t − 1 ) + t + 1 1 + t − 1 1 − ( t + 1 ) 2 1 + ( t − 1 ) 2 1 ) + C = = 2 1 ( ln t + 1 t − 1 ) + ( t 2 − 1 ) t − ( t 2 − 1 ) 2 2 t + C t = x 4 − 4 x 2 t = \frac {\sqrt {x ^ {4} - 4}}{x ^ {2}} t = x 2 x 4 − 4 ∫ x 5 x 4 − 4 d x = 1 2 ( ln x 4 − 4 x 2 − 1 x 4 − 4 x 2 + 1 ) + x 4 − 4 x 2 ( x 4 − 4 x 4 − 1 ) − 2 x 4 − 4 x 2 ( x 4 − 4 x 4 − 1 ) 2 + C = \int x ^ {5} \sqrt {x ^ {4} - 4} d x = \frac {1}{2} \left(\ln \frac {\frac {\sqrt {x ^ {4} - 4}}{x ^ {2}} - 1}{\frac {\sqrt {x ^ {4} - 4}}{x ^ {2}} + 1}\right) + \frac {\frac {\sqrt {x ^ {4} - 4}}{x ^ {2}}}{\left(\frac {x ^ {4} - 4}{x ^ {4}} - 1\right)} - \frac {2 \frac {\sqrt {x ^ {4} - 4}}{x ^ {2}}}{\left(\frac {x ^ {4} - 4}{x ^ {4}} - 1\right) ^ {2}} + C = ∫ x 5 x 4 − 4 d x = 2 1 ( ln x 2 x 4 − 4 + 1 x 2 x 4 − 4 − 1 ) + ( x 4 x 4 − 4 − 1 ) x 2 x 4 − 4 − ( x 4 x 4 − 4 − 1 ) 2 2 x 2 x 4 − 4 + C = = 1 2 ln x 4 − 4 − x 2 x 4 − 4 + x 2 − x 2 x 4 − 4 8 + x 6 x 4 − 4 4 + C , C is an arbitrary real constant. = \frac {1}{2} \ln \frac {\sqrt {x ^ {4} - 4} - x ^ {2}}{\sqrt {x ^ {4} - 4} + x ^ {2}} - \frac {x ^ {2} \sqrt {x ^ {4} - 4}}{8} + \frac {x ^ {6} \sqrt {x ^ {4} - 4}}{4} + C, C \text{ is an arbitrary real constant.} = 2 1 ln x 4 − 4 + x 2 x 4 − 4 − x 2 − 8 x 2 x 4 − 4 + 4 x 6 x 4 − 4 + C , C is an arbitrary real constant.
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#340153 on Dec 2023