What is $f (x) = \int \frac{x}{\sqrt{x^{2} + 1}} d x$ $f(x) = int x/sqrt(x^2+1) dx$ if $f (2) = 3$ $f(2) = 3$ ?

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What is $f (x) = \int \frac{x}{\sqrt{x^{2} + 1}} d x$ $f(x) = int x/sqrt(x^2+1) dx$ if $f (2) = 3$ $f(2) = 3$ ?

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Answer 1 · 2016-03-20T20:51:36

$f (x) = \sqrt{x^{2} + 1} + 3 - \sqrt{5}$ $f(x)=sqrt(x^2+1)+3-sqrt5$

Explanation:

To integrate:

Let $u = x^{2} + 1$ $u=x^2+1$ so $d u = 2 x d x$ $du=2xdx$ .

This gives us:

$f (x) = \frac{1}{2} \int \frac{2 x}{\sqrt{x^{2} + 1}} d x = \frac{1}{2} \int \frac{1}{\sqrt{u}} d u = \frac{1}{2} \int u^{- \frac{1}{2}} d u$ $f(x)=1/2int(2x)/sqrt(x^2+1)dx=1/2int1/sqrtudu=1/2intu^(-1/2)du$

Now, we integrate using the rule: $\int u^{n} d u = \frac{u^{n + 1}}{n + 1}$ $intu^ndu=u^(n+1)/(n+1)$

So, we have

$f (x) = \frac{1}{2} \frac{u^{\frac{1}{2}}}{\frac{1}{2}} + C = \frac{1}{2} \sqrt{u} \cdot 2 + C = \sqrt{u} + C$ $f(x)=1/2u^(1/2)/(1/2)+C=1/2sqrtu*2+C=sqrtu+C$

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

Using the original condition $f (2) = 3$ $f(2)=3$ , we have

$3 = \sqrt{2^{2} + 1} + C$ $3=sqrt(2^2+1)+C$

$3 = \sqrt{5} + C$ $3=sqrt5+C$

$C = 3 - \sqrt{5}$ $C=3-sqrt5$

So, substituting this in, we see that

$f (x) = \sqrt{x^{2} + 1} + 3 - \sqrt{5}$ $f(x)=sqrt(x^2+1)+3-sqrt5$

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What is $f (x) = \int \frac{x}{\sqrt{x^{2} + 1}} d x$ $f(x) = int x/sqrt(x^2+1) dx$ if $f (2) = 3$ $f(2) = 3$ ?

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Explanation:

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What is f(x) = int x/sqrt(x^2+1) dxf(x)=∫x√x2+1dxf(x) = int x/sqrt(x^2+1) dx if f(2) = 3 f(2)=3f(2) = 3 ?

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What is $f (x) = \int \frac{x}{\sqrt{x^{2} + 1}} d x$ $f(x) = int x/sqrt(x^2+1) dx$ if $f (2) = 3$ $f(2) = 3$ ?