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  • Evaluate the definite integrals in Exercises 25 to 33. 29.

Evaluate the definite integrals in Exercises 25 to 33.

    Q29.    \int_0^1\frac{dx}{\sqrt{1+x} -\sqrt x}

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best_answer

\int_0^1\frac{dx}{\sqrt{1+x} -\sqrt x}

First, let's get rid of the square roots from the denominator,

\\=\int_0^1\frac{dx}{\sqrt{1+x} -\sqrt x}*\frac{\sqrt{1+x} +\sqrt x}{\sqrt{1+x} +\sqrt x}

\\=\int_0^1\frac{\sqrt{1+x}+\sqrt{x}}{{1+x} -x}dx

\\=\int_0^1({\sqrt{1+x}+\sqrt{x}})dx

\\=\int_0^1({\sqrt{1+x})dx+\int_0^1({\sqrt{x}})dx

\\=\int_0^1(1+x)^\frac{1}{2}dx+\int_0^1x^\frac{1}{2}dx

\\=\left [ \frac{2}{3}(1+x)^{\frac{3}{2}} \right ]_0^1+\left [ \frac{2}{3}(x)^{\frac{3}{2}} \right ]_0^1

\\=\left [ \frac{2}{3}(1+1)^{\frac{3}{2}} \right ]-\left [ \frac{2}{3} \right ]+\left [ \frac{2}{3}(1)^{\frac{3}{2}} \right ]-\left [ 0 \right ]

\\=\frac{4\sqrt{2}}{3}

Posted by

Pankaj Sanodiya

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