What is the antiderivative of # (ln(x+1)/(x^2))#?

Answer 1

#I = -ln(x+1)/x + ln|x| - ln|x+1| + c#

#I = intln(x+1)/x^2dx#
Let's say #u = ln(x+1)# so #du = 1/(x+1)#, and #dv = 1/x^2# so #v = -1/x#
#I = -ln(x+1)/x + intdx/(x(x+1))#

The latter integral can only be solved with parcial fractions, so we assume there is a sum of fractions

#a/x + b/(x+1) = 1/(x(x+1))#
Where #a# and #b# are constants, that is also to say
#(a(x+1) + bx)/(x(x+1)) = 1/(x(x+1))#
Note that it means, that, for any value of #x#
#a(x+1) + bx = 1#
So if we assume #x = 0#,
#a(0+1) + b*0 = 1# #a = 1#
And if we assume #x = -1#
#a(-1+1) -b = 1# #-b = 1# #b = -1#

So, back to the first integral we have

#I = -ln(x+1)/x + int(1/x-1/(x+1))dx# #I = -ln(x+1)/x + intdx/x -intdx/(x+1)#

From there, it's an easy integral

#I = -ln(x+1)/x + ln|x| - ln|x+1| + c#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

The antiderivative of ( \frac{\ln(x+1)}{x^2} \The antiderivative of ( \frac{\ln(x+1)}{x^2} ) with respectThe antiderivative of ( \frac{\ln(x+1)}{x^2} ) cannotThe antiderivative of ( \frac{\ln(x+1)}{x^2} ) with respect toThe antiderivative of ( \frac{\ln(x+1)}{x^2} ) cannot beThe antiderivative of ( \frac{\ln(x+1)}{x^2} ) with respect to (The antiderivative of ( \frac{\ln(x+1)}{x^2} ) cannot be expressedThe antiderivative of ( \frac{\ln(x+1)}{x^2} ) with respect to ( xThe antiderivative of ( \frac{\ln(x+1)}{x^2} ) cannot be expressed inThe antiderivative of ( \frac{\ln(x+1)}{x^2} ) with respect to ( x \The antiderivative of ( \frac{\ln(x+1)}{x^2} ) cannot be expressed in termsThe antiderivative of ( \frac{\ln(x+1)}{x^2} ) with respect to ( x )The antiderivative of ( \frac{\ln(x+1)}{x^2} ) cannot be expressed in terms ofThe antiderivative of ( \frac{\ln(x+1)}{x^2} ) with respect to ( x ) cannotThe antiderivative of ( \frac{\ln(x+1)}{x^2} ) cannot be expressed in terms of elementaryThe antiderivative of ( \frac{\ln(x+1)}{x^2} ) with respect to ( x ) cannot beThe antiderivative of ( \frac{\ln(x+1)}{x^2} ) cannot be expressed in terms of elementary functions.The antiderivative of ( \frac{\ln(x+1)}{x^2} ) with respect to ( x ) cannot be expressed in termsThe antiderivative of ( \frac{\ln(x+1)}{x^2} ) cannot be expressed in terms of elementary functions.The antiderivative of ( \frac{\ln(x+1)}{x^2} ) with respect to ( x ) cannot be expressed in terms of elementary functions.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7