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Ierofanga [76]
3 years ago
6

Please help me to solve this ∫ 1/(x + √x) dx

Mathematics
1 answer:
skad [1K]3 years ago
6 0

Answer:

\boxed{2ln (\sqrt{x}  + 1 ) + c}

Step-by-step explanation:

\int\limits {\frac{1}{x + \sqrt{x} } } \, dx

= \int\limits {\frac{1}{\sqrt{x} (\sqrt{x}  + 1 )} } \, dx

= \int\limits {\frac{2}{(\sqrt{x}  + 1 )} } \, d(\sqrt{x}  + 1 )

= 2ln (\sqrt{x}  + 1 ) + c

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Step-by-step explanation:

The true statements are:

Over the interval [2, 4], the local minimum is –8.

Over the interval [3, 5], the local minimum is –8.

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Over the interval [1, 3], the local minimum is 0

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Over the interval [2, 4], the local minimum is –8.

This statement is true because the given minimum point is(3.4, -8). Thus the  local minimum is -8 which is true

Over the interval [3, 5], the local minimum is –8.

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Over the interval [1, 4], the local maximum is 0.

Look at the graph. The maximum point given is (2,0). Thus this statement is true because local maximum is 0.

Over the interval [3, 5], the local maximum is 0.

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