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

• azizlwl
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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.

azizlwl said:
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.

azizlwl and hutchphd
azizlwl said:
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".

azizlwl
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
$$\int \mathrm{d} x \frac{x^2+2x}{(x+1)^2}=x+\frac{1}{x+1}+C.$$

Why am I getting the wrong answer for the integral of (x^2 + 2x)/(x + 1)^2?

One common reason is not simplifying the integrand properly before integrating. The expression can often be simplified or decomposed into partial fractions, which makes the integration process more straightforward.

How do I simplify the integrand (x^2 + 2x)/(x + 1)^2?

To simplify, you can perform polynomial long division on (x^2 + 2x) by (x + 1). This will help you rewrite the integrand in a form that's easier to integrate, typically resulting in a simpler rational function or polynomial plus a simpler fraction.

Can I use partial fraction decomposition for (x^2 + 2x)/(x + 1)^2?

Yes, partial fraction decomposition can be used. However, since (x + 1)^2 is already a factor in the denominator, you might first rewrite the numerator to a form that makes it easier to decompose. In this case, polynomial long division might be more straightforward.

What is the correct approach to integrate (x^2 + 2x)/(x + 1)^2?

First, use polynomial long division to rewrite (x^2 + 2x)/(x + 1)^2. The integrand can be simplified to a form like A/(x + 1) + B/(x + 1)^2, where A and B are constants. Then, integrate each term separately.

What is the final integral of (x^2 + 2x)/(x + 1)^2 after simplification?

After simplifying (x^2 + 2x)/(x + 1)^2, you might get an expression like (x - 1)/(x + 1) + 1/(x + 1). Integrating this, you get ∫(x - 1)/(x + 1) dx + ∫1/(x + 1) dx, which results in x - 2ln|x + 1| + C, where C is the constant of integration.

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