I'm getting the wrong answer for the Indefinite Integral of: (x^2+2x)/(x+1)^2

[azizlwl]

Homework Statement

(x^2+2x)/(x+1)^2
Ans: x^2/(x+1)

Relevant Equations

Integral

((x+1)^2 -1)/(x+1)^2 dx
1-1/(x+1)^2 dx
Let u=x+1
1-1/u^2 du
u+1/u +c
(u^2+1)/u +c
Not as answer given in the book.

 

[cbarker1]

Homework Statement: (x^2+2x)/(x+1)^2
Ans: x^2/(x+1)
Relevant Equations: Integral

((x+1)^2 -1)/(x+1)^2 dx
1-1/(x+1)^2 dx
Let u=x+1
1-1/u^2 du
u+1/u +c
(u^2+1)/u +c
Not as answer given in the book.

You may have an equivalent answer to the book. There are so many equivalent answers to one indefinite integrals. I recommend you to take the derivative of your answer to see if it is your integrand.

 

[Mark44]

Homework Statement: (x^2+2x)/(x+1)^2
Ans: x^2/(x+1)
Relevant Equations: Integral

((x+1)^2 -1)/(x+1)^2 dx
1-1/(x+1)^2 dx
Let u=x+1
1-1/u^2 du
u+1/u +c
(u^2+1)/u +c
Not as answer given in the book.

1. You didn't "undo" your substitution. When you use this technique of integration, you should always rewrite your answer in terms of the original variable, not the substitution variable.
2. Your answer, reverting back to the original variable x, is ##x + 1 + \frac 1 {x + 1} + C##.
If I subtract the answer shown in the book from your answer, I get a constant. If two people work an indefinite integral by different methods, they can often come up with different-appearing solutions. If the two solutions differ only by a constant, then differentiating each solution will result in the given integrand.

One more thing: you've been a member here for over ten years. If you're going to post questions about mathematics, do yourself a favor and learn a bit about how to post using LaTeX. There's a link to our tutorial in the lower left corner of the input pane, "LaTeX Guide".

 

[vanhees71]

Hm, it's perhaps easier to first write
$$\frac{x^2+2x}{(x+1)^2}=\frac{(x+1)^2-1}{(x+1)^2}=1-\frac{1}{(x+1)^2},$$
which you can immediately integrate
$$\int \mathrm{d} x \frac{x^2+2x}{(x+1)^2}=x+\frac{1}{x+1}+C.$$