Evaluate the indefinite integral $\int \frac{2x\,dx}{x^2 +2x+1}$

Question

Here is my attempt:

$$ \frac{2x}{x^2 +2x+1}= \frac{2x}{(x+1)^2 } = \frac{2}{x+1}-\frac{2}{(x+1)^2 }$$

Then I tried to integrate it,I got $2\ln(x+1)+\frac{2}{x+1}+C$ as my answer. Am I right? please correct me if I'm wrong.

Answer 1

Hint The factorization $(x + 1)^2$ of the denominator of the integrand suggests that we can rewrite the integral using the substitution $u := x + 1$, $du = dx$: $$\int \frac{2 (u - 1)}{u^2} du = 2 \left(\int \frac{du}{u} - \int \frac{du}{u^2} \right) .$$

Answer 2

Your answer is almost correct, but you should put modulus sign in your log's argument as it must be positive i.e your answer should be; $$2\ln|x+1|+\frac{2}{x+1}+C$$

You can try one more method too.

$$\int \frac{2x}{x^2+2x+1}dx$$ Try to create derivative of denominator in numerator $$\int \frac{2x+2-2}{x^2+2x+1}dx$$ $$\int \frac{2x+2}{x^2+2x+1}dx-2\int \frac{1}{x^2+2x+1}dx$$

$$\int \frac{2x+2}{x^2+2x+1}dx-2\int \frac{1}{(x+1)^2}dx$$ For first integral simply take $x^2+2x+1=t$, then $(2x+2)dx=dt$, hence you will get: $$\ln(x^2+2x+1)+\frac{2}{x+1}+C$$ $$\ln(x+1)^2+\frac{2}{x+1}+C$$ $$2\ln|x+1|+\frac{2}{x+1}+C$$

Answer 3

Add and remove $2$ in the numerator, and you will obtain:

$$\frac{2x+2-2}{x^2+2x+1} = \frac{2x+2}{x^2+2x+1} - \frac{2}{x^2+2x+1}$$

Now you see that the numerator in the first fraction is nothing but the derivative of the denominator, id est you have a function of the form $\frac{f'(x)}{f(x)}$. This hen integrated is simply

$$\ln(x^2+2x+1)$$

Now the other term

$$2\int\frac{1}{x^2 + 2x + 1}\ \text{d}x = 2\int\frac{1}{(x+1)^2}\ \text{d}x = -\frac{2}{(x+1)}$$

Obtained by simply using $y = (x+1)$.

Final result:

$$\ln(x^2+2x+1) -\frac{2}{(x+1)}$$

The logarithm can be written also as

$$\ln((x+1)^2) = 2\ln(x+1)$$

Answer 4

$$\int\frac{2x}{x^2+2x+1}\text{d}x$$

let $u=x^2+2x+1$ and $\frac{du}{dx}=2x+2$ and now $dx=\frac{du}{2(x+1)}$

$$\int\frac{2x}{u}\cdot \frac{du}{2(x+1)}$$

Now since $x^2+2x+1=(x+1)^2$, therefore $u^\frac{1}{2}=x+1$ then $u^\frac{1}{2}-1=x$

$$\int\frac{u^\frac{1}{2}-1}{u^\frac{3}{2}}du$$

$$\int\frac{u^\frac{1}{2}}{u^\frac{3}{2}}du-\int\frac{1}{u^\frac{3}{2}}du$$

$$\int\frac{1}{u}du--2u^{-\frac{1}{2}}$$

$$2\cdot ln|x+1|+\frac{2}{x+1}+C$$

Answer 5

$$=\int\frac{2x+2}{x^2+2x+1}dx-2\int\frac{dx}{x^2+2x+1}$$

set $t=x^2+2x+1$ and $dt=(2x+2)dx$

$$=\int\frac{1}{t}-2\int\frac{dx}{(x+1)^2}dx$$

Set $\nu=x+1$ and $d\nu=dx$

$$=\int\frac{1}{t}-2\int\frac{1}{\nu ^2}d\nu=\ln|t|+\frac{2}{\nu}+\mathcal C=\color{red}{\ln|x^2+2x+1|+\frac{2}{x+1}+\mathcal C}$$