Question

integral of (1+sqrt x)^8?

Answer 1

It can not be calculated.

Answer 2

This is under integral strategy subtopic. Any idea where to begin anyone?

Answer 3

Funny thing, here...

\(\int \left(\dfrac{1}{1+\sqrt{x}}\right)^{8}\;dx = \int \left(1+\sqrt{x}\right)^{-8}\;dx\)

Take a wild guess. It MIGHT look something like this...

\(\dfrac{1}{-7}\left(1+\sqrt{x}\right)^{-7}\)

Suppose we take the derivative of this and see how close we are?

\(\dfrac{d}{dx}\dfrac{1}{-7}\left(1+\sqrt{x}\right)^{-7} = \dfrac{1}{2\sqrt{x}(1+\sqrt{x})^{8}}\)

Oh, we're SO CLOSE!!!! Let's just play like the differences are constants.

\(\dfrac{d}{dx}\dfrac{2\sqrt{x}}{-7}\left(1+\sqrt{x}\right)^{-7} = \dfrac{1}{(1+\sqrt{x})^{8}} - \dfrac{1}{7\sqrt{x}(1+\sqrt{x})^{7}}\)

Again, this is SO CLOSE to what we need. If only that last term had an easy antiderivative!!

Oh, wait, \(\int \dfrac{1}{7\sqrt{x}(1+\sqrt{x})^{7}}\;dx = \dfrac{-1}{21(1+\sqrt{x})^{6}} + C\)

And thus we see...

\(\int \dfrac{1}{(1+\sqrt{x})^{8}}\;dx = -\dfrac{2}{7}\dfrac{\sqrt{x}}{(1+\sqrt{x})^{7}} - \dfrac{1}{21}\dfrac{1}{(1+\sqrt{x})^{6}} + C\)

That can be simplified a little.

Answer 4

The "Where to begin" question is answered.

Repeated here for your convenience: "Take a wild guess. It MIGHT look something like this..."