How do you evaluate the integral #int x^2/(16+x^2)#?

Question

William Abdalla · Accepted Answer

I got:

#x - 4arctan(x/4) + C#

Using a trick:

#int (16 + x^2 - 16)/(16 + x^2)dx#

#= int dx - 16int 1/(16 + x^2)dx#

#= int dx - int 1/(1 + (x/4)^2)dx#

Now, for the second integral only, we would let #u = x/4# to find that #du = 1/4dx#. Thus, we multiply by #4# to maintain the integrand:

#= int dx - 4int 1/4 1/(1 + (x/4)^2)dx#

#= int dx - 4int 1/(1 + u^2)dx#

Now, if we recognize that #d/(dx)[arctanx] = 1/(1+x^2)#, then we have:

#=> int dx - 4arctanu#

#= color(blue)(x - 4arctan(x/4) + C)#

Isabella Ackerman · Answer

undefined To evaluate the integral $\int \frac{x^2}{16+x^2} \, dx$, we use the method of partial fractions.

First, we write the integrand as $\frac{x^2}{16+x^2} = \frac{A}{4+x} + \frac{B}{4-x}$ where A and B are constants to be determined.

Next, we find common denominators and equate coefficients to get $x^2 = A(4-x) + B(4+x)$.

Solving for A and B, we find $A = 1$ and $B = 1$.

Now, we rewrite the integral as $\int \frac{1}{4+x} + \frac{1}{4-x} \, dx$.

Integrating term by term, we get $\int \frac{1}{4+x} \, dx + \int \frac{1}{4-x} \, dx = \ln|4+x| - \ln|4-x| + C$, where C is the constant of integration.

So, $\int \frac{x^2}{16+x^2} \, dx = \ln|\frac{4+x}{4-x}| + C$, where C is the constant of integration.