Question

Help with num 5 please.​

Answer 1

Let's apply the derivative to both sides with respect to x. We'll use the chain rule.

`y = \ln\left(x + √(x^2+a^2)\right)\\\\\\(dy)/(dx) = (d)/(dx)\left[\ln\left(x + √(x^2+a^2)\right)\right]\\\\\\(dy)/(dx) = (1)/(x+√(x^2+a^2))*(d)/(dx)\left[x + √(x^2+a^2)\right]\\\\\\(dy)/(dx) = (1)/(x+(x^2+a^2)^(1/2))*\left(1 + 2x*(1)/(2)(x^2+a^2)^(-1/2)\right)\\\\\\(dy)/(dx) = (1)/(x+(x^2+a^2)^(1/2))*\left(1 + x*(1)/((x^2+a^2)^(1/2))\right)\\\\\\(dy)/(dx) = (1)/(x+W)*\left(1 + x*(1)/(W)\right)\\\\\\`

where W = (x^2+a^2)^(1/2) = sqrt(x^2+a^2)

Let's simplify that a bit.

`(dy)/(dx) = (1)/(x+W)*\left(1 + x*(1)/(W)\right)\\\\\\(dy)/(dx) = (1)/(x+W)*\left((W)/(W) + (x)/(W)\right)\\\\\\(dy)/(dx) = (1)/(x+W)*(W+x)/(W)\\\\\\(dy)/(dx) = (1)/(W)\\\\\\(dy)/(dx) = (1)/(√(x^2+a^2))\\\\\\`

This concludes the first part of what your teacher wants.

-----------------------

Now onto the second part.

In this case, a = 3 so a^2 = 3^2 = 9.

Recall that if
`g(x) = (d)/(dx)\left[f(x)\right]`

then
`\displaystyle \int g(x)dx = f(x)+C`

We can say that f(x) is the antiderivative of g(x). The C is some constant.

So,

`\displaystyle \displaystyle \int (1)/(√(x^2+a^2))dx = \ln\left(x + √(x^2+a^2)\right)+C\\\\\\\displaystyle \displaystyle \int (1)/(√(x^2+3^2))dx = \ln\left(x + √(x^2+3^2)\right)+C\\\\\\\displaystyle \displaystyle \int (1)/(√(x^2+9))dx = \ln\left(x + √(x^2+9)\right)+C\\\\\\`

Now let g(x) = ln(x + sqrt(x^2+9) ) + C

Then compute

• g(0) = ln(3)+C
• g(4) = ln(9) = ln(3^2) = 2*ln(3)+C

Therefore,

`\displaystyle \displaystyle \int (1)/(√(x^2+9))dx = \ln\left(x + √(x^2+9)\right)+C\\\\\\\displaystyle \displaystyle \int (1)/(√(x^2+9))dx = g(x)\\\\\\\displaystyle \displaystyle \int_(0)^(4) (1)/(√(x^2+9))dx = g(4) - g(0)\\\\\\\displaystyle \displaystyle \int_(0)^(4) (1)/(√(x^2+9))dx = (2\ln(3)+C)-(\ln(3)+C)\\\\\\\displaystyle \displaystyle \int_(0)^(4) (1)/(√(x^2+9))dx = \ln(3)\\\\\\`

---------------------------------------

Extra info:

Interestingly, WolframAlpha says that the result is
`\sinh^(-1)\left((4)/(3)\right)`, but we can rewrite that into ln(3) because inverse hyperbolic sine is defined as

`\sinh^(-1)(x) = \ln\left(x + √(x^2+1)\right)`

which is the function your teacher gave you, but now a = 1.

If you plugged x = 4/3 into the hyperbolic sine definition above, then you should get
`\sinh^(-1)\left((4)/(3)\right) = \ln\left(3\right)`