How do solve #∫x tan^-1x dx#, given that #d/dx tan^-1x = 1/(1+x^2)# ?

Question

Samantha Adkison · Accepted Answer

#((x^2+1)tan^-1(x)-x)/2+C#

#I=intxtan^-1(x)dx#

We will use integration by parts. Integration by parts takes the form #intudv=uv-intvdu#.

Since we cannot integrate #tan^-1(x)# well, let #u=tan^-1(x)# (notice you've also been provided with the derivative of #tan^-1(x)# in the question). Thus, let #dv# be the remaining portion: #dv=xdx#.

Differentiating #u# and integrating #dv#:

Then:

#I=uv-intvdu=x^2/2tan^-1(x)-intx^2/2(1/(1+x^2))dx#

Simplifying slightly:

#I=x^2/2tan^-1(x)-1/2intx^2/(1+x^2)dx#

Rewrite the integrand as follows, or perform polynomial long division on #x^2/(1+x^2)#:

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

#I=x^2/2tan^-1(x)-1/2int(1+x^2)/(1+x^2)dx-1/2int1/(1+x^2)dx#

#I=x^2/2tan^-1(x)-1/2intdx-1/2int(-1)/(1+x^2)dx#

Note that #intdx=x#. Furthermore, since #d/dxtan^-1(x)=1/(1+x^2)#, we see that #int1/(1+x^2)dx=tan^-1(x)#.

#I=x^2/2tan^-1(x)-1/2x+1/2tan^-1(x)#

Which we may write as:

#I=(x^2tan^-1(x)-x+tan^-1(x))/2#

Rearranging and adding the constant of integration:

#I=((x^2+1)tan^-1(x)-x)/2+C#

Emma Aldridge · Answer

undefined To solve ∫x tan^-1x dx, use integration by parts, where u = tan^-1x and dv = x dx.

This gives:
du = (1/(1+x^2)) dx
v = (1/2)x^2

Apply integration by parts formula:

∫u dv = uv - ∫v du

Substitute the values of u, v, du, and dv:

= (x * tan^-1x * (1/2)x^2) - ∫((1/2)x^2 * (1/(1+x^2))) dx

= (x^3/2) * tan^-1x - (1/2) * ∫(x^2/(1+x^2)) dx

Now, let's solve the remaining integral by using a substitution method.

Let u = x^2 + 1, du = 2x dx
Thus, dx = (1/2) du

Substituting, we get:

= (x^3/2) * tan^-1x - (1/4) * ∫(1/u) du

= (x^3/2) * tan^-1x - (1/4) * ln|u| + C

Finally, substitute back x^2 + 1 for u:

= (x^3/2) * tan^-1x - (1/4) * ln|x^2 + 1| + C