Evaluate the integral? : #int x^2/(x^2+1)^2 dx#

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William Abdalla · Accepted Answer

Option #1

# int \ x^2/(x^2+1)^2 dx = 1/2arctanx - x/(2(x^2+1)) + C #

Let us denote the given integral by #I#: # I = int \ x^2/(x^2+1)^2 dx # We can simply the fractional integrand with a slight manipulation: # I = int \ (x^2+1-1)/(x^2+1)^2 \ dx # # \ \ = int \ (x^2+1)/(x^2+1)^2 - (1)/(x^2+1)^2 \ dx # # \ \ = int \ 1/(x^2+1) \ dx - int \ (1)/(x^2+1)^2 \ dx # # \ \ = arctanx - int \ (1)/(x^2+1)^2 \ dx + C# For this next integral we can use a substitution: Let # tan theta = x => x^2 +1 = tan^2 theta+1 = sec^2 theta # # dx/(d theta)=sec^2 theta # Substituting we see that: # int \ (1)/(x^2+1)^2 \ dx = int \ (1)/(sec^2 theta)^2\ sec^2 theta \ d theta # # " " = int \ (1)/(sec^2 theta)^2\ sec^2 theta \ d theta # # " " = int \ (1)/(sec^2 theta) \ d theta # # " " = int \ cos^2 theta \ d theta # # " " = int \ 1/2(1+cos2theta) \ d theta # # " " = 1/2 \ int \ 1+cos2theta \ d theta # # " " = 1/2 (theta + 1/2sin2theta) # # " " = 1/2 (theta + sinthetacostheta) # # " " = 1/2 (theta + sinthetacostheta * costheta/costheta)# # " " = 1/2 (theta + sintheta/costheta cos^2theta)# # " " = 1/2 (theta + tan theta/sec^2theta)# # " " = 1/2 (arctanx + x/(x^2+1) )# Combining this with the earlier result gives: # I = arctanx - 1/2 (arctanx + x/(x^2+1) ) + C # # \ \ = 1/2arctanx - x/(2(x^2+1)) + C #

Luke Alvarez · Answer

undefined To evaluate the integral $\int \frac{x^2}{{(x^2+1)}^2} \, dx$, you can use the method of partial fractions. After performing the partial fraction decomposition, the integral can be expressed as the sum of two simpler integrals, which can then be evaluated using standard integration techniques. The result is:

$\int \frac{x^2}{{(x^2+1)}^2} \, dx = \frac{1}{2}\left(\ln|x^2+1| - \frac{x^2}{x^2+1}\right) + C$

where $C$ is the constant of integration.

Isabella Ackerman · Answer

undefined To evaluate the integral $\int \frac{x^2}{(x^2+1)^2} \,dx$, we use a substitution and partial fractions if necessary. However, for this integral, a direct approach can be taken recognizing a pattern:

Let $I = \int \frac{x^2}{(x^2+1)^2} \,dx$.

Notice that the derivative of $x^2+1$ is $2x$, which suggests a substitution or an adjustment to the integrand to match this derivative. However, a straightforward approach involves noticing a relationship to the derivative of the denominator itself.

The integral can be approached by adding and subtracting 1 in the numerator:

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

The first term on the right is easier to evaluate:

1. $\int \frac{x^2 + 1}{(x^2+1)^2} \,dx = \int \frac{1}{x^2 + 1} \,dx = \tan^{-1}(x) + C_1$

For the second term, you can use a standard integral:

2. $\int \frac{1}{(x^2+1)^2} \,dx$

This integral is recognized as a standard form, whose result is $\frac{1}{2}\left(\frac{x}{x^2 + 1} + \tan^{-1}(x)\right) + C_2$.

Putting it all together:

$$I = \tan^{-1}(x) - \frac{1}{2}\left(\frac{x}{x^2 + 1} + \tan^{-1}(x)\right) + C$$

$$I = \frac{1}{2}\tan^{-1}(x) - \frac{1}{2}\cdot\frac{x}{x^2 + 1} + C$$

So, the integral evaluates to:

$$\int \frac{x^2}{(x^2+1)^2} \,dx = \frac{1}{2}\tan^{-1}(x) - \frac{1}{2}\cdot\frac{x}{x^2 + 1} + C$$

Where $C$ is the constant of integration.