How do you integrate #f(x)=lnx/(1+x^2)# using the quotient rule?

Question

Isabella Ackerman · Accepted Answer

There is no quotient rule for integrating.

There is a quotient rule for differentiating, perhaps that is what was meant. #d/dx(u/v) = (u'v-uv')/v^2# In this case #u = lnx#, so #u' = 1/x# and #v = 1+x^2# #d/dx((lnx)/(1+x^2)) = (1/x(1+x^2)-(lnx)(2x))/(1+x^2)^2# # = (1+x^2-2x^2lnx)/(x(1+x^2)^2)#

Emma Aldridge · Answer

undefined To integrate $ f(x) = \frac{\ln x}{1 + x^2} $ using the quotient rule, we first express the logarithmic function as a product of functions:

$$ \ln x = \ln x \cdot 1 $$

Then, we apply integration by parts, which states:

$$ \int u \, dv = uv - \int v \, du $$

Letting $ u = \ln x $ and $ dv = \frac{1}{1 + x^2} \, dx $, we find $ du = \frac{1}{x} \, dx $ and $ v = \arctan x $. Plugging these into the integration by parts formula, we get:

$$ \int \frac{\ln x}{1 + x^2} \, dx = \ln x \cdot \arctan x - \int \arctan x \cdot \frac{1}{x} \, dx $$

The remaining integral $ \int \arctan x \cdot \frac{1}{x} \, dx $ can be solved using integration by substitution. Letting $ u = \arctan x $, we have $ du = \frac{1}{1 + x^2} \, dx $. This gives us:

$$ \int \arctan x \cdot \frac{1}{x} \, dx = \int u \, du = \frac{u^2}{2} + C = \frac{(\arctan x)^2}{2} + C $$

Therefore, the final result is:

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

where $ C $ is the constant of integration.