Evaluate the following integral. \int \frac{x}{\sqrt{x-4}}dx

Evaluate the following integral. \int \frac{x}{\sqrt{x-4}}dx

Question
Applications of integrals
asked 2021-02-14
Evaluate the following integral.
\(\displaystyle\int{\frac{{{x}}}{{\sqrt{{{x}-{4}}}}}}{\left.{d}{x}\right.}\)

Answers (1)

2021-02-15
Step 1
Consider the integrals,
\(\displaystyle\int{\frac{{{x}}}{{\sqrt{{{x}-{4}}}}}}{\left.{d}{x}\right.}\)
Suppose that,
\(\displaystyle\sqrt{{{x}-{4}}}={t}\)
Differentiating with respect to "x"
\(\displaystyle{\frac{{{1}}}{{{2}\sqrt{{{x}-{4}}}}}}{\left.{d}{x}\right.}={\left.{d}{t}\right.}\)
\(\displaystyle{\frac{{{1}}}{{\sqrt{{{x}-{4}}}}}}{\left.{d}{x}\right.}={2}{\left.{d}{t}\right.}\)
Step 2
Substitute all value in given integrals,
\(\displaystyle\int{\frac{{{x}}}{{\sqrt{{{x}-{4}}}}}}{\left.{d}{x}\right.}=\int{\left({t}^{{{2}}}+{4}\right)}{2}\ {\left.{d}{t}\right.}\)
\(\displaystyle={2}\int{\left({t}^{{{2}}}+{4}\right)}{\left.{d}{t}\right.}\)
\(\displaystyle={2}{\left[{\frac{{{t}^{{{3}}}}}{{{3}}}}+{4}{t}\right]}+{C}\)
\(\displaystyle={\frac{{{2}{t}^{{{3}}}}}{{{3}}}}+{8}{t}+{C}\)
\(\displaystyle={\frac{{{2}}}{{{3}}}}{\left({x}-{4}\right)}^{{{\frac{{{3}}}{{{2}}}}}}+{8}\sqrt{{{\left({x}-{4}\right)}}}+{C}\)
0

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