How do you evaluate the integral $\int \frac{1}{x \sqrt{x^{2} + 1}}$ $int 1/(xsqrt(x^2+1))$ ?

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How do you evaluate the integral $\int \frac{1}{x \sqrt{x^{2} + 1}}$ $int 1/(xsqrt(x^2+1))$ ?

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Answer 1 · 2017-01-11T01:56:22

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

Explanation:

Substitute:

$t = \sqrt{1 + x^{2}}$ $t = sqrt(1+x^2)$

$d t = \frac{x d x}{\sqrt{1 + x^{2}}}$ $dt = (xdx)/sqrt(1+x^2)$

We have:

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

And as:

$t = \sqrt{1 + x^{2}} \Rightarrow t^{2} - 1 = x^{2}$ $t = sqrt(1+x^2) => t^2-1 = x^2$

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

This can be solved by partial fractions:

$\frac{1}{t^{2} - 1} = \frac{1}{(t - 1) (t + 1)} = \frac{A}{t - 1} + \frac{B}{t + 1}$ $1/(t^2-1) = 1/((t-1)(t+1)) = A/(t-1) + B/(t+1)$

$A (t + 1) + B (t - 1) = 1$ $A(t+1)+B(t-1) = 1$

$(A + B) t + (A - B) = 1$ $(A+B) t + (A-B) = 1$

$A + B = 0 \Rightarrow A = - B$ $A+B = 0 => A=-B$

$A - B = 1 \Rightarrow 2 A = 1 \Rightarrow A = \frac{1}{2} \Rightarrow B = - \frac{1}{2}$ $A-B = 1 => 2A = 1 => A=1/2 => B =-1/2$

So:

$\frac{1}{t^{2} - 1} = \frac{1}{2} (\frac{1}{t - 1} - \frac{1}{t + 1})$ $1/(t^2-1) = 1/2(1/(t-1)-1/(t+1))$

and:

$\int \frac{d x}{x \sqrt{1 + x^{2}}} = \frac{1}{2} \int \frac{d t}{t - 1} - \frac{1}{2} \int \frac{d t}{t + 1} = \frac{1}{2} ln | t - 1 | - \frac{1}{2} ln | t + 1 | + C$ $int (dx)/(xsqrt(1+x^2)) = 1/2 int (dt)/(t-1) - 1/2 int (dt)/(t+1)= 1/2 ln abs (t-1) - 1/2 ln abs (t+1)+C$

Substituting back $x$ $x$ :

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

where we have dropped the absolute value, because the arguments of the logarithms are always positive. Applying the properties of logarithms this can also be written as:

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

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How do you evaluate the integral $\int \frac{1}{x \sqrt{x^{2} + 1}}$ $int 1/(xsqrt(x^2+1))$ ?

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