Question Number 5265 by Kasih last updated on 03/May/16
$$\int\:\frac{{x}}{\mathrm{1}+{x}}\:{dx} \\ $$
Answered by Yozzii last updated on 03/May/16
$$\frac{{x}}{\mathrm{1}+{x}}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{x}}=\frac{\mathrm{1}+{x}−\mathrm{1}}{\mathrm{1}+{x}}=\frac{{x}}{\mathrm{1}+{x}} \\ $$$$\Rightarrow\int\frac{{x}}{\mathrm{1}+{x}}{dx}=\int\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{x}}\right){dx}={x}−{ln}\mid{x}+\mathrm{1}\mid+{C} \\ $$$$ \\ $$
Commented by Kasih last updated on 09/May/16
$${can}\:{it}\:{use}\:{integration}\:{by}\:{parts}? \\ $$
Commented by 1771727373 last updated on 14/May/16
$${no}\:{you}\:{cant} \\ $$