Question #5f659

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Question #5f659

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Answer 1 · 2017-03-14T00:44:10

$2 \sqrt{1 + ln x} + ln ∣ ∣ ∣ \frac{\sqrt{1 + ln x} - 1}{\sqrt{1 + ln x} + 1} ∣ ∣ ∣ + C$ $2sqrt(1+lnx)+lnabs((sqrt(1+lnx)-1)/(sqrt(1+lnx)+1))+C$

Explanation:

Use the substitution $u = 1 + ln x$ $u=1+lnx$ . This implies that $d u = \frac{1}{x} d x$ $du=1/xdx$ and $ln x = u - 1$ $lnx=u-1$ . Then:

$\int \frac{\sqrt{1 + ln x}}{x ln x} d x = \int \frac{\sqrt{u}}{u - 1} d u$ $intsqrt(1+lnx)/(xlnx)dx=intsqrtu/(u-1)du$

Now let $v = \sqrt{u}$ $v=sqrtu$ . This implies that $v^{2} = u$ $v^2=u$ , so $(2 v) d v = d u$ $(2v)dv=du$ .

$= \int \frac{v}{v^{2} - 1} (2 v) d v = 2 \int \frac{v^{2}}{v^{2} - 1} d v$ $=intv/(v^2-1)(2v)dv=2intv^2/(v^2-1)dv$

Rewriting the integrand as $\frac{v^{2} - 1 + 1}{v^{2} - 1} = 1 + \frac{1}{v^{2} - 1}$ $(v^2-1+1)/(v^2-1)=1+1/(v^2-1)$ :

$= 2 \int (1 + \frac{1}{v^{2} - 1}) d v$ $=2int(1+1/(v^2-1))dv$

We can perform partial fraction decomposition on $\frac{1}{v^{2} - 1} = \frac{1}{(v + 1) (v - 1)}$ $1/(v^2-1)=1/((v+1)(v-1))$ :

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

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

Letting $v = 1$ $v=1$ reveals that $B = \frac{1}{2}$ $B=1/2$ and letting $v = - 1$ $v=-1$ reveals that $A = - \frac{1}{2}$ $A=-1/2$ . Thus

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

Then the integral becomes:

$= 2 \int (1 + \frac{1}{2 (v - 1)} - \frac{1}{2 (v + 1)}) d v$ $=2int(1+1/(2(v-1))-1/(2(v+1)))dv$

$= 2 \int d v + \int \frac{d v}{v - 1} - \int \frac{d v}{v + 1}$ $=2intdv+int(dv)/(v-1)-int(dv)/(v+1)$

These are simple integrals to find:

$= 2 v + ln (| v - 1 |) - ln (| v + 1 |) + C$ $=2v+ln(abs(v-1))-ln(abs(v+1))+C$

$= 2 \sqrt{u} + ln ∣ ∣ ∣ \frac{\sqrt{u} - 1}{\sqrt{u} + 1} ∣ ∣ ∣ + C$ $=2sqrtu+lnabs((sqrtu-1)/(sqrtu+1))+C$

$= 2 \sqrt{1 + ln x} + ln ∣ ∣ ∣ \frac{\sqrt{1 + ln x} - 1}{\sqrt{1 + ln x} + 1} ∣ ∣ ∣ + C$ $=2sqrt(1+lnx)+lnabs((sqrt(1+lnx)-1)/(sqrt(1+lnx)+1))+C$

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Question #5f659

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