Question

How do you find the indefinite integral of `int sqrt [(144x^2) + 196] / x dx`?

Answer 1

`2sqrt(36x^2+49)+7ln(abs(sqrt(36x^2+49)-7))-7ln(sqrt(36x^2+49)+7)+C`

Explanation:

Some manipulation will be involved. First, simplify the square root.

`intsqrt(144x^2+196)/xdx=2intsqrt(36x^2+49)/xdx`

Multiply the integrand by `sqrt(36x^2+49)/sqrt(36x^2+49)`.

`=2int(36x^2+49)/(xsqrt(36x^2+49))dx`

Let `u=sqrt(36x^2+49)`, implying that `du=(36x)/sqrt(36x^2+49)dx`.

Since we already have the `sqrt(36x^2+49)` needed for `du` in the denominator, we just need `36x` in the numerator, so multiply the integrand by `(36x)/(36x)`.

`=2int(36x(36x^2+49))/(36x^2sqrt(36x^2+49))dx`

We can rearrange this so that we have `du=(36x)/sqrt(36x^2+49)dx` present in the integrand.

`=2int(36x^2+49)/(36x^2)((36x)/sqrt(36x^2+49))dx`

For the remaining part of the portion that is not part of the `du` expression, we can rewrite these in terms of `u`.

In the numerator, we have `36x^2+49`. Since `u=sqrt(36x^2+49)`, we know that `u^2=36x^2+49`.

In the denominator, since `u^2=36x^2+49`, we know that `u^2-49=36x^2`.

Combining these with what we've already found, we see the integrand equals:

`=2intu^2/(u^2-49)du`

To integrate this, we must split up this fraction. You can do so through polynomial long division, or by writing `u^2` as `u^2-49+49` and then splitting up the fraction from there.

`=2int(u^2-49+49)/(u^2-49)du=2int1+49/(u^2-49)du`

The integrand can be split up and we can easily integrate the first portion (the constant of integration will be added at the end):

`=2intdu+98int1/(u^2-49)du=2u+98int1/(u^2-49)du`

To integrate the remaining piece, write `1/(u^2-49)=1/((u+7)(u-7))` in its partial fractions.

`1/((u+7)(u-7))=A/(u+7)+B/(u-7)`

`1=A(u-7)+B(u+7)`

Setting `u=7` reveals that `1=14B=>B=1/14`.

Setting `u=-7` reveals that `1=-14A=>A=-1/14`.

Thus, `1/(u^2-49)=1/(14(u-7))-1/(14(u+7))`. Substituting this into our partially solved integral expression, we see that:

`2u+98int1/(u^2-49)du=2u+98int1/(14(u-7))du+98int1/(14(u+7))du`

`=2u+7int1/(u-7)du-7int1/(u+7)du`

Both of these are in the form `int1/vdv`, where `v=u-7` and `v=u+7`, respectively. They are both simply integrated using natural logarithms:

`=2u+7ln(abs(u-7))-7ln(abs(u+7))+C`

Using `u=sqrt(36x^2+49)`, we obtain the antiderivative:

`=2sqrt(36x^2+49)+7ln(abs(sqrt(36x^2+49)-7))-7ln(sqrt(36x^2+49)+7)+C`

The absolute value brackets are not needed for `ln(sqrt(36x^2+49)+7)` since `sqrt(36x^2+49)+7>0` for all values of `x`.