Evaluate $\int \frac{1}{1+x}\, dx$

Question

I forgot about integrals so I need some help in this problem $\displaystyle\int \frac{1}{1+x}\, dx$ please.

Answer 1

Hint: Use the $u$-substitution $u=1+x$ then $\mathrm du=\mathrm dx$ so the integral becomes $$\int\dfrac1u\,\mathrm du.$$

Answer 2

Your integral: $$\int \frac{1}{1+x} \ \text dx$$ Use the u-substitution $u=1+x$. Then $\text du = \text dx$ $$\int \frac 1u \ \text du$$ The integral of $\frac{1}{u}$ is $\ln|u| + C$ $$\int \frac 1u \ du = \ln|u| + C$$ Reverse the substitution. $$\color{green}{\int \frac{1}{1+x} \ \text dx=\ln\left|1+x \right|+C}$$ Hope I helped!

Answer 3

A primitive is log|1+x|, because of the rule that a primitive of u'/u is ln|u|.

Because this solution is not defined in at the point x=-1, the general solution is log(1+x) + C_1 when x > -1 and log(-(1+x)) + C_2 when x < -1.