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

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

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Answer 1 · 2015-05-13T13:41:36

This is hard...

$\int \frac{1}{\sqrt{1 + \sqrt{1 + x^{2}}}} d x$ $int1/sqrt(1+sqrt(1+x^2))dx$

We need to delete all this square !!

For this let's :

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

$x = \sqrt{u^{2} - 1}$ $x = sqrt(u^2-1)$

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

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

Integral become :

$\int \frac{\sqrt{1 + x^{2}}}{x \cdot \sqrt{1 + \sqrt{1 + x^{2}}}} d u$ $intsqrt(1+x^2)/(x*sqrt(1+sqrt(1+x^2)))du$

$\int \frac{u}{\sqrt{u^{2} - 1} \cdot \sqrt{u + 1}} d u$ $intu/(sqrt(u^2-1)*sqrt(u+1))du$

Multiply numerator and denominator by $\sqrt{u - 1}$ $sqrt(u-1)$ and don't forget that $(u + 1) (u - 1) = u^{2} - 1$ $(u+1)(u-1) = u^2-1$ so we have :

$\frac{u \sqrt{u - 1}}{u^{2} - 1}$ $(usqrt(u-1))/(u^2-1)$

Let's $w = \sqrt{u - 1}$ $w = sqrt(u-1)$

$d w = \frac{1}{2 \sqrt{u - 1}} d u$ $dw = 1/(2sqrt(u-1))du$

$d u = 2 \sqrt{u - 1} d w$ $du = 2sqrt(u-1)dw$

$2 \int \frac{u \sqrt{u - 1}}{u^{2} - 1} \sqrt{u - 1} d w$ $2int(usqrt(u-1))/(u^2-1)sqrt(u-1)dw$

$w^{2} + 1 = u$ $w^2+1=u$

${(w^{2} + 1)}^{2} = u^{2}$ $(w^2+1)^2=u^2$

$2 \int \frac{(w^{2} + 1) w^{2}}{{(w^{2} + 1)}^{2} - 1} d w$ $2int((w^2+1)w^2)/((w^2+1)^2-1)dw$

$2 \int \frac{w^{4} + w^{2}}{w^{4} + 2 w^{2}} d w$ $2int(w^4+w^2)/(w^4+2w^2)dw$

$2 \int \frac{w^{2} (w^{2} + 1)}{w^{2} (w^{2} + 2)} d w$ $2int(w^2(w^2+1))/(w^2(w^2+2))dw$

$2 \int \frac{w^{2} + 1}{w^{2} + 2} d w$ $2int(w^2+1)/(w^2+2)dw$

$2 \int \frac{w^{2} + 2 - 1}{w^{2} + 2} d w$ $2int(w^2+2-1)/(w^2+2)dw$

$2 \int 1 d w - 2 \int \frac{1}{w^{2} + 2} d w$ $2int1dw-2int1/(w^2+2)dw$

$2 \int 1 d w - \int \frac{1}{\frac{1}{2} w^{2} + 1} d w$ $2int1dw-int1/(1/2w^2+1)dw$

Let's $t = \frac{1}{\sqrt{2}} w$ $t =1/sqrt(2)w$

$t^{2} = \frac{1}{2} w^{2}$ $t^2=1/2w^2$

$d t = \frac{1}{\sqrt{2}} d w$ $dt = 1/sqrt(2)dw$

$[2 w] - \sqrt{2} \int \frac{1}{t^{2} + 1} d t$ $[2w]- sqrt(2)int1/(t^2+1)dt$

$[2 w] - [\sqrt{2} arctan (t)] + C$ $[2w]-[sqrt(2)arctan(t)]+C$

Substitute back...

$[2 w] - [\sqrt{2} arctan (\frac{1}{\sqrt{2}} w)] + C$ $[2w]-[sqrt(2)arctan(1/sqrt(2)w)]+C$

$[2 \sqrt{u - 1}] - [\sqrt{2} arctan (\frac{1}{\sqrt{2}} \sqrt{u - 1})] + C$ $[2sqrt(u-1)]-[sqrt(2)arctan(1/sqrt(2)sqrt(u-1))]+C$

$[2 \sqrt{\sqrt{x^{2} + 1} - 1}] - [\sqrt{2} arctan (\frac{1}{\sqrt{2}} \sqrt{\sqrt{x^{2} + 1} - 1})] + C$ $[2sqrt(sqrt(x^2+1)-1)]-[sqrt(2)arctan(1/sqrt(2)sqrt(sqrt(x^2+1)-1))]+C$

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

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