How do you evaluate the integral $\int \frac{x}{\sqrt{4 x + 1}}$ $int x/sqrt(4x+1)$ ?

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How do you evaluate the integral $\int \frac{x}{\sqrt{4 x + 1}}$ $int x/sqrt(4x+1)$ ?

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Answer 1 · 2017-04-10T20:19:25

$\int \frac{x}{\sqrt{4 x + 1}} d x = \frac{2 x - 1}{12} \sqrt{4 x + 1} + C$ $int x/sqrt(4x+1) dx = (2x-1)/12sqrt(4x+1) +C$

Explanation:

Note that:

$\frac{1}{\sqrt{4 x + 1}} = \frac{1}{2} \frac{d}{d x} \sqrt{4 x + 1}$ $1/sqrt(4x+1) = 1/2 d/dx sqrt(4x+1)$

So we can write the integral as:

$\int \frac{x}{\sqrt{4 x + 1}} d x = \frac{1}{2} \int x d (\sqrt{4 x + 1})$ $int x/sqrt(4x+1) dx = 1/2 int x d(sqrt(4x+1))$

and integrate by parts:

$\int \frac{x}{\sqrt{4 x + 1}} d x = \frac{1}{2} x \sqrt{4 x + 1} - \frac{1}{2} \int \sqrt{4 x + 1} d x$ $int x/sqrt(4x+1) dx = 1/2xsqrt(4x+1) -1/2 int sqrt(4x+1)dx$

The resulting integral can be resolved directly using the power rule:

$\int \frac{x}{\sqrt{4 x + 1}} d x = \frac{1}{2} x \sqrt{4 x + 1} - \frac{1}{8} \int {(4 x + 1)}^{\frac{1}{2}} d (4 x + 1)$ $int x/sqrt(4x+1) dx = 1/2xsqrt(4x+1) -1/8 int (4x+1)^(1/2)d(4x+1)$

$\int \frac{x}{\sqrt{4 x + 1}} d x = \frac{1}{2} x \sqrt{4 x + 1} - \frac{1}{8} \frac{{(4 x + 1)}^{\frac{3}{2}}}{\frac{3}{2}} + C$ $int x/sqrt(4x+1) dx = 1/2xsqrt(4x+1) -1/8 (4x+1)^(3/2)/(3/2)+C$

and simplifying:

$\int \frac{x}{\sqrt{4 x + 1}} d x = \frac{1}{2} x \sqrt{4 x + 1} - \frac{1}{12} (4 x + 1) \sqrt{4 x + 1} + C$ $int x/sqrt(4x+1) dx = 1/2xsqrt(4x+1) -1/12 (4x+1)sqrt(4x+1)+C$

$\int \frac{x}{\sqrt{4 x + 1}} d x = (\frac{1}{2} x - \frac{1}{3} x - \frac{1}{12}) \sqrt{4 x + 1} + C$ $int x/sqrt(4x+1) dx = (1/2x-1/3x-1/12)sqrt(4x+1) +C$

$\int \frac{x}{\sqrt{4 x + 1}} d x = \frac{2 x - 1}{12} \sqrt{4 x + 1} + C$ $int x/sqrt(4x+1) dx = (2x-1)/12sqrt(4x+1) +C$

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How do you evaluate the integral $\int \frac{x}{\sqrt{4 x + 1}}$ $int x/sqrt(4x+1)$ ?

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