Find # I = int \ lnx/x^2 \ dx #?

Answer 1

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

We seek:

# I = int \ lnx/x^2 \ dx #

We can apply integration by by parts:

Let # { (u,=lnx, => (du)/dx,=1/x), ((dv)/dx,=1/x^2, => v,=-1/x ) :}#

Then plugging into the IBP formula:

# int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx #

gives us

# int \ (lnx)(1/x^2) \ dx = (lnx)(-1/x) - int \ (-1/x)(1/x) \ dx #
# :. int \ (lnx)/x^2 \ dx = -(lnx)/x + int \ 1/x^2 \ dx #
# :. int \ (lnx)/x^2 \ dx = -(lnx)/x - 1/x + C #
# :. int \ (lnx)/x^2 \ dx = -(lnx+1)/x + 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

To solve the integral ( I = \int \frac{\ln x}{x^2} , dx ), we can use integration by parts.

Let ( u = \ln x ) and ( dv = \frac{1}{x^2} , dx ).

Differentiating ( u ) gives ( du = \frac{1}{x} , dx ) and integrating ( dv ) gives ( v = -\frac{1}{x} ).

Now apply the integration by parts formula:

[ I = \int u , dv = uv - \int v , du ]

Substitute the values of ( u ), ( v ), ( du ), and ( dv ) into the formula:

[ I = -\ln x \cdot \frac{1}{x} - \int \left(-\frac{1}{x}\right) \cdot \frac{1}{x} , dx ]

Simplify and integrate the remaining integral:

[ I = -\frac{\ln x}{x} + \int \frac{1}{x^2} , dx ]

[ I = -\frac{\ln x}{x} - \frac{1}{x} + C ]

Thus, the integral is ( I = -\frac{\ln x}{x} - \frac{1}{x} + C ), where ( C ) is the constant of integration.

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