What is $f (x) = \int x - \sqrt{x^{2} + 1} d x$ $f(x) = int x-sqrt(x^2+1) dx$ if $f (2) = 7$ $f(2) = 7$ ?

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What is $f (x) = \int x - \sqrt{x^{2} + 1} d x$ $f(x) = int x-sqrt(x^2+1) dx$ if $f (2) = 7$ $f(2) = 7$ ?

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Answer 1 · 2017-01-15T15:20:30

$f (x) = \frac{1}{2} x^{2} - \frac{1}{2} x \sqrt{x^{2} + 1} - \frac{1}{2} ln ∣ ∣ \sqrt{x^{2} + 1} + x ∣ ∣ + 5 + \sqrt{5} + \frac{1}{2} ln (\sqrt{5} + 2)$ $f(x)=1/2x^2-1/2xsqrt(x^2+1)-1/2lnabs(sqrt(x^2+1)+x)+5+sqrt5+1/2ln(sqrt5+2)$

Explanation:

$f (x) = \int x - \sqrt{x^{2} + 1} . d x$ $f(x)=intx-sqrt(x^2+1)color(white).dx$

Split up the integral:

$f (x) = \int x . d x - \int \sqrt{x^{2} + 1} . d x$ $f(x)=intxcolor(white).dx-intsqrt(x^2+1)color(white).dx$

The first is easily done:

$f (x) = \frac{1}{2} x^{2} - \int \sqrt{x^{2} + 1} . d x$ $f(x)=1/2x^2-intsqrt(x^2+1)color(white).dx$

Let $J = \int \sqrt{x^{2} + 1} . d x$ $J=intsqrt(x^2+1)color(white).dx$ .

Solve the remaining integral with the substitution $x = tan θ$ $x=tantheta$ . This implies that $d x = {sec}^{2} θ . d θ$ $dx=sec^2thetacolor(white).d theta$ . So:

$J = \int \sqrt{{tan}^{2} θ + 1} ({sec}^{2} θ . d θ)$ $J=intsqrt(tan^2theta+1)(sec^2thetacolor(white).d theta)$

Note that $1 + {tan}^{2} θ = {sec}^{2} θ$ $1+tan^2theta=sec^2theta$ so $\sqrt{1 + {tan}^{2} θ} = sec θ$ $sqrt(1+tan^2theta)=sectheta$ :

$J = \int {sec}^{3} θ . d θ$ $J=intsec^3thetacolor(white).d theta$

This can be solved through integration by parts. Let:

${(u=sectheta" "=>" "du=secthetatanthetacolor(white).d theta),(dv=sec^2thetacolor(white).d theta" "=>" "v=tantheta):}$

Therefore:

$J=secthetatantheta-intsecthetatan^2thetacolor(white).d theta$

Letting $tan^2theta=sec^2theta-1$ :

$J=secthetatantheta-intsectheta(sec^2theta-1)color(white).d theta$

$J=secthetatantheta-intsec^3thetacolor(white).d theta+intsecthetacolor(white).d theta$

Note that the original integral is back on the right side, so we can replace it with $J$ . Also note that $intsecthetacolor(white).d theta=lnabs(sectheta+tantheta)$ , which is a well-known integral.

$J=secthetatantheta-J+lnabs(sectheta+tantheta)$

Adding $J$ to both sides of the equation then dividing by $2$ to solve for $J$ again gives:

$J=1/2secthetatantheta+1/2lnabs(sectheta+tantheta)$

Now returning to the original integral:

$f(x)=1/2x^2-J$

$f(x)=1/2x^2-1/2secthetatantheta-1/2lnabs(sectheta+tantheta)$

Now we can return to $x$ from $theta$ . Our original substitution was $x=tantheta$ . This also implies that $sectheta=sqrt(tan^2theta+1)=sqrt(x^2+1)$ . Hence:

$f(x)=1/2x^2-1/2xsqrt(x^2+1)-1/2lnabs(sqrt(x^2+1)+x)+C$

Solve for $C$ using the original condition $f(2)=7$ :

$7=1/2(2^2)-1/2(2)sqrt(2^2+1)-1/2lnabs(sqrt(2^2+1)+2)+C$

$7=2-sqrt5-1/2ln(sqrt5+2)+C$

So:

$C=5+sqrt5+1/2ln(sqrt5+2)$

Finally:

$f(x)=1/2x^2-1/2xsqrt(x^2+1)-1/2lnabs(sqrt(x^2+1)+x)+5+sqrt5+1/2ln(sqrt5+2)$

Answer 2 · 2017-01-15T15:44:10

$f(x) = 1/2x^2 -1/2xsqrt(1 + x^2) - 1/2ln|x + sqrt(1 + x^2)| + 6.51$

Explanation:

ATTENTION: LONG ANSWER AHEAD!!

$f(x) = intxdx - intsqrt(x^2 + 1)dx$

We are going to need trig substitution to solve $intsqrt(x^2 + 1)dx$ .

Let $x = tantheta$ . Then $dx= sec^2theta d theta$ .

$intsqrt((tan theta)^2 + 1)sec^2theta d theta$

$intsqrt(sec^2theta) sec^2theta d theta$

$intsec^3theta d theta$

We can use integration by parts to attack this problem. The formula is $int(udv) = uv - int(vdu)$ .

Let $dv = sec^2theta d theta$ . Then $v = tantheta$ . Let $u = sectheta$ . Then $du = secthetatantheta d theta$ .

$intsec^3thetad theta = tan thetasectheta - int(tanthetasecthetatantheta d theta)$

$intsec^3thetad theta = tan thetasectheta - inttan^2thetasecthetad theta$

$intsec^3thetad theta = tanthetasectheta - int(sec^2theta - 1)sec theta d theta$

$intsec^3thetad theta = tan thetasectheta - intsec^3thetad theta + intsectheta d theta$

We are now faced with a problem. In the integral, we are left with $intsec^3thetad theta$ , which is what we were having troubles with at first. The solution is to add $intsec^3thetad theta$ to both sides.

$intsec^3thetad theta + intsec^3thetad theta= tanthetasectheta - intsec^3thetad theta + intsec^3thetad theta + intsectheta d theta$

$2intsec^3thetad theta = tanthetasectheta + intsectheta d theta$

$2intsec^3thetad theta = tan thetasectheta + ln|tantheta + sectheta| + C$

$intsec^3theta d theta = 1/2tanthetasectheta +1/2ln|tantheta + sectheta| + C$

Reverse the substitutions by drawing a triangle and finding the correct ratios.

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$intsqrt(x^2 + 1)dx = 1/2xsqrt(1 + x^2) + 1/2ln|x + sqrt(1 + x^2)| + C$

Now put this together with the $int(x)dx= 1/2x^2$ to get:

$f(x) = 1/2x^2 - (1/2xsqrt(1 + x^2) + 1/2ln|x + sqrt(1 + x^2)|) + C$

$f(x) = 1/2x^2 -1/2xsqrt(1 + x^2) - 1/2ln|x + sqrt(1 + x^2)| + C$

All we have to do now is solve for $C$ :

$7= 1/2(2)^2 - 1/2(2)sqrt(1 + 2^2) - 1/2ln|2 + sqrt(1 + 2^2)| + C$

$7 = 2 - sqrt(5) + 1/2ln|2 + sqrt(5)| + C$

$5 + sqrt(5) - 1/2ln(2 + sqrt(5) = C$

Use a calculator to find the approximation $6.51$ .

Hence, $f(x) = 1/2x^2 -1/2xsqrt(1 + x^2) - 1/2ln|x + sqrt(1 + x^2)| + 6.51$

Hopefully this helps!

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What is $f (x) = \int x - \sqrt{x^{2} + 1} d x$ $f(x) = int x-sqrt(x^2+1) dx$ if $f (2) = 7$ $f(2) = 7$ ?

2 Answers

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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) = 7 f(2)=7f(2) = 7 ?

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What is $f (x) = \int x - \sqrt{x^{2} + 1} d x$ $f(x) = int x-sqrt(x^2+1) dx$ if $f (2) = 7$ $f(2) = 7$ ?