Question

Can someone help me with the antiderivative of this function.

Answer 1

\[\int\limits_{}^{}\frac{ x+1 }{ \sqrt{x} - 1 } \]

Answer 2

@freckles @FibonacciChick666

Answer 3

is it justthe squareroot of x on the bottom?

Answer 4

Yes

Answer 5

ok, I recommend multiplying by the conjugate to see if it gets you something nicer

Answer 6

This is the answer. To be honest I don't even know how to approach this one.

Answer 7

oh wow lol

Answer 8

yes lol

Answer 9

\[\int\limits_{}^{}\frac{x+1}{\sqrt{x}-1}dx = \int\limits_{}^{}\frac{x}{\sqrt{x}-1}dx + \int\limits_{}^{}\frac{1}{\sqrt{x}-1}dx\]
and then integrate by parts?

Answer 10

yah gives u right sndwer

Answer 11

x/sqrx-1 will take some work

Answer 12

I lied; it's a substitution problem

Answer 13

You split the integral into 2 like before
For each integral, let:\[u = \sqrt(x)-1\]
Which means \[du = \frac{1}{2\sqrt{x}}dx\]
Which means\[dx = 2\sqrt{x} * du\]

Answer 14

\(\frac{2 x^{3/2}}{3}+x+4 \sqrt{x}+4 \log \left(1-\sqrt{x}\right)\)

Answer 15

is it easier if we make it look like this:\[\int \frac{(x+1)(\sqrt{x} +1)}{x-1}\]

Answer 16

then we can use polynomial division to annihilate part of it

Answer 17

\[dx = 2(u+1)du\]

Answer 18

For the integrand \(\frac{x+1}{\sqrt{x}-1}\), substitute \(u=\sqrt{x}\) and \(du=\frac{1}{2 \sqrt{x}} \, dx\):
\(= 2 \int \frac{u \left(u^2+1\right)}{u-1} \, du\)

For the integrand \(\frac{u \left(u^2+1\right)}{u-1}\), do long division:
\(= 2 \int \left(u^2+u+\frac{2}{u-1}+2\right) \, du\)

Answer 19

Integrate the sum term by term and factor out constants:
\(= 2 \int u^2 \, du+4 \int 1 \, du+4 \int \frac{1}{u-1} \, du+2 \int u \, du\)

Answer 20

For the first integral, we substitute in and get:
\[\int\limits_{}^{}\frac{(u+1)^2}{u}(2[u+1]du)\]
which is equal to
\[2\int\limits_{}^{}\frac{(u+1)^3}{u}du\]
This integral should be straight forward

Answer 21

This is a serious question. :/

Answer 22

For the second integral susbstitution, we get
\[\int\limits_{}^{}\frac{1}{u}*dx\]
substitute the dx from before, and you should get another straightforward integral

Answer 23

I like alchemista's substitution better

Answer 24

I like it too

Answer 25

Everyone deserves a medal for this one