Question

integral of sqrt(x^2+2x)

Answer 1

hint:
one way in order to solve such integral, is to make this substitution:
\[\Large \sqrt {{x^2} + 2x} = x + t\]
where \(t\) is the new variable

Answer 2

after that substitution, we get:
\[\Large \int {dx} \sqrt {{x^2} + 2x} = \int {dt\frac{{{{\left( {2t - {t^2}} \right)}^2}}}{{4{{\left( {1 - t} \right)}^3}}}} \]

Answer 3

after integrating that I get this which is incorrect :(

Answer 4

Just kidding! Plugged in (x^2+2x)^.5-x!!! Thanks!!!!

Answer 5

\[I=\int\limits \sqrt{x^2+2x}*1 dx=\sqrt{x^2+2x}*x-\int\limits \frac{ 2x+2 }{2\sqrt{x^2+2x} }*x dx+c\]
\[=x \sqrt{x^2+2x}-\int\limits \frac{ x^2+2x-x }{ \sqrt{x^2+2x} }dx+c\]
\[=x \sqrt{x^2+2x}-\int\limits \frac{ x^2+2x }{ \sqrt{x^2+2x} }dx+\int\limits \frac{ x }{ \sqrt{x^2+2x} }dx+c\]
\[=x \sqrt{x^2+2x}-\int\limits \sqrt{x^2+2x}dx+\frac{ 1 }{ 2 }\int\limits \frac{ 2x+2-2 }{ \sqrt{x^2+2x} }dx+c\]
\[=x \sqrt{x^2+2x}-I+\frac{ 1 }{ 2 }\int\limits \left( x^2+2x \right)^{\frac{ -1 }{ 2 }}\left( 2x+2 \right)dx-\int\limits \frac{ dx }{\sqrt{x^2+2x} }+c\]
\[2 I=?\]