Question

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

Answer 1

`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)=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)`

Answer 2

`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.

`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!