Question

integral

Answer 1

stuck on
\[I=\int\limits \frac{x+\sqrt{x+1}}{x+2}dx\]

Answer 2

@mathmath333

Answer 3

try substitute \(u=\sqrt{x+1}\)

Answer 4

I tried something like this
\[I=\int\limits dx-2 \int\limits \frac{dx}{x+2}+\int\limits \frac{\sqrt{x+1}}{x+2}dx\]

Answer 5

this one also looks good

Answer 6

\[u=\sqrt{x+1}\]\[dx=2\times u \times du\]
third integral,
\[2\int\limits \frac{u^2}{u^2+1}du\]

Answer 7

yes solve it

Answer 8

ill give it a try

Answer 9

u can apply long division

Answer 10

\(\large \begin{align} \color{black}{\dfrac{u^2}{u^2+1}\hspace{.33em}\\~\\
=\dfrac{u^2+1-1}{u^2+1}\hspace{.33em}\\~\\
=1-\dfrac{1}{u^2+1}\hspace{.33em}\\~\\
}\end{align}\)

Answer 11

\[I=x-2\log|x+1|+2\sqrt{x+1}-2\tan^-1\sqrt{x+1}+C\]
?

Answer 12

thats correct

Answer 13

Answer in book is
\[I=(x+1)+2\sqrt{x+1}-2\log|x+2|-2\tan^-1\sqrt{x+1}+C\]

Answer 14

answer books has an extra one which is a typo

Answer 15

http://www.wolframalpha.com/input/?i=I%3D%5Cint%5Climits+%5Cfrac%7Bx%2B%5Csqrt%7Bx%2B1%7D%7D%7Bx%2B2%7Ddx