Ortiz Dickson
06/20/2023 · Senior High School
Evaluate the integral by completing a square \( I=\int \frac{1}{\sqrt{-x^{2}+4 x-3}} d x \) A. None of the options B. \( \sin ^{-1}(x-2)+x+C \) C. \( -\sin ^{-1}(x-1)+C \) D. \( \cos ^{-1}(x-2)+C \) E. \( \sin ^{-1}(x-2)+C \)
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Tutor-Verified Answer
Step-by-step Solution
Calculate the integral \( \int \frac{1}{\sqrt{-x^{2}+4x-3}} dx \).
Evaluate the integral by following steps:
- step0: Evaluate using substitution:
\(\int \frac{1}{\sqrt{-x^{2}+4x-3}} dx\)
- step1: Evaluate the power:
\(\int \frac{1}{\left(-x^{2}+4x-3\right)^{\frac{1}{2}}} dx\)
- step2: Rewrite the expression:
\(\int \frac{1}{\left(1-\left(x-2\right)^{2}\right)^{\frac{1}{2}}} dx\)
- step3: Use the substitution \(t=x-2\) to transform the integral\(:\)
\(\int \frac{1}{\left(1-t^{2}\right)^{\frac{1}{2}}} dt\)
- step4: Evaluate the integral:
\(\arcsin\left(t\right)\)
- step5: Substitute back:
\(\arcsin\left(x-2\right)\)
- step6: Add the constant of integral C:
\(\arcsin\left(x-2\right) + C, C \in \mathbb{R}\)
The integral \( I = \int \frac{1}{\sqrt{-x^{2}+4x-3}} dx \) is equal to \( \arcsin(x-2) + C \), where \( C \) is an arbitrary constant.
Therefore, the correct answer is E. \( \sin ^{-1}(x-2)+C \).
Quick Answer
The correct answer is E. \( \sin ^{-1}(x-2)+C \).
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