What is #f(x) = int x-sqrt(x^2+1) dx# if #f(2) = 7 #?

#f(x)=intx-sqrt(x^2+1)color(white).dx#

Split up the integral:

#f(x)=intxcolor(white).dx-intsqrt(x^2+1)color(white).dx#

The first is easily done:

#f(x)=1/2x^2-intsqrt(x^2+1)color(white).dx#

Let #J=intsqrt(x^2+1)color(white).dx#.

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

#J=intsqrt(tan^2theta+1)(sec^2thetacolor(white).d theta)#

Note that #1+tan^2theta=sec^2theta# so #sqrt(1+tan^2theta)=sectheta#:

#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)#

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.

#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!