Q. \[ \int \frac{\ln x}{x^2}\, dx \]

Q: What is \int \frac{\ln x}{x^2}\,dx?

Use integration by parts. Let u=\ln x and dv=x^{-2}\,dx. Then du=\frac{1}{x}\,dx and v=-\frac{1}{x}. Now apply the integration by parts formula: \int u\,dv=uv-\int v\,du So: \int \frac{\ln x}{x^2}\,dx=\ln x\left(-\frac{1}{x}\right)-\int\left(-\frac{1}{x}\right)\frac{1}{x}\,dx Simplify: \int \frac{\ln x}{x^2}\,dx=-\frac{\ln x}{x}+\int\frac{1}{x^2}\,dx Since \frac{1}{x^2}=x^{-2}, we have: \int x^{-2}\,dx=-x^{-1}=-\frac{1}{x} Therefore: \int \frac{\ln x}{x^2}\,dx=-\frac{\ln x}{x}-\frac{1}{x}+C Factor the numerator: \int \frac{\ln x}{x^2}\,dx=-\frac{\ln x+1}{x}+C Final answer: \int \frac{\ln x}{x^2}\,dx=-\frac{\ln x+1}{x}+C

Q: How do you derive -\frac{\ln x+1}{x}?

Start from integration by parts: \int \frac{\ln x}{x^2}\,dx=\ln x\left(-\frac{1}{x}\right)-\int\left(-\frac{1}{x}\right)\frac{1}{x}\,dx Simplify the first term: \ln x\left(-\frac{1}{x}\right)=-\frac{\ln x}{x} Simplify the integral: -\int\left(-\frac{1}{x}\right)\frac{1}{x}\,dx=\int\frac{1}{x^2}\,dx Now integrate: \int\frac{1}{x^2}\,dx=\int x^{-2}\,dx=-x^{-1}=-\frac{1}{x} So: \int \frac{\ln x}{x^2}\,dx=-\frac{\ln x}{x}-\frac{1}{x}+C Combine the terms: -\frac{\ln x}{x}-\frac{1}{x}=-\frac{\ln x+1}{x} Therefore: \int \frac{\ln x}{x^2}\,dx=-\frac{\ln x+1}{x}+C

Q: What is \int \frac{\ln x}{x^2}\,dx for x\gt0?

For real-valued logarithms, \ln x requires x\gt0. The antiderivative is still: \int \frac{\ln x}{x^2}\,dx=-\frac{\ln x+1}{x}+C So, for x\gt0: \int \frac{\ln x}{x^2}\,dx=-\frac{\ln x+1}{x}+C

Q: Is there a substitution method instead of integration by parts?

Let t=\ln x. Then x=e^t, and dx=e^t\,dt. Rewrite the integrand: \frac{\ln x}{x^2}\,dx=\frac{t}{e^{2t}}\cdot e^t\,dt Simplify: \frac{t}{e^{2t}}\cdot e^t\,dt=te^{-t}\,dt So the integral becomes: \int te^{-t}\,dt This integral still needs integration by parts. It gives the same final result: \int \frac{\ln x}{x^2}\,dx=-\frac{\ln x+1}{x}+C

Q: What is \int \frac{\ln x+1}{x^2}\,dx?

Split the fraction into two parts: \frac{\ln x+1}{x^2}=\frac{\ln x}{x^2}+\frac{1}{x^2} We know: \int \frac{\ln x}{x^2}\,dx=-\frac{\ln x+1}{x}+C Also: \int \frac{1}{x^2}\,dx=-\frac{1}{x} So: \int \frac{\ln x+1}{x^2}\,dx=-\frac{\ln x+1}{x}-\frac{1}{x}+C Simplify: \int \frac{\ln x+1}{x^2}\,dx=-\frac{\ln x+2}{x}+C Final answer: \int \frac{\ln x+1}{x^2}\,dx=-\frac{\ln x+2}{x}+C

Q: Can you check by differentiation that -\frac{\ln x+1}{x}+C is correct?

Let: f(x)=-\frac{\ln x+1}{x} Rewrite it using a negative exponent: f(x)=-(\ln x+1)x^{-1} Differentiate using the product rule. f'(x)=-\left[\frac{1}{x}\cdot x^{-1}+(\ln x+1)(-1)x^{-2}\right] Distribute the negative sign: f'(x)=-\frac{1}{x^2}+(\ln x+1)x^{-2} Rewrite x^{-2} as \frac{1}{x^2}. f'(x)=-\frac{1}{x^2}+\frac{\ln x+1}{x^2} Combine the fractions: f'(x)=\frac{-1+\ln x+1}{x^2} Simplify the numerator: f'(x)=\frac{\ln x}{x^2} Since the derivative gives the original integrand, the antiderivative is correct.

Answer

We want

\[
\int \frac{\ln x}{x^2}\,dx.
\]

Use integration by parts. Let \(u=\ln x\), so \(du=\frac{1}{x}\,dx\). Let \(dv=x^{-2}\,dx\), so \(v=\int x^{-2}\,dx=-\frac{1}{x}\).

Then

\[
\int \frac{\ln x}{x^2}\,dx=u\,v-\int v\,du
= \ln x\left(-\frac{1}{x}\right)-\int\left(-\frac{1}{x}\right)\left(\frac{1}{x}\right)\,dx.
\]

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

Final result:

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

Detailed Explanation

We want to compute the indefinite integral

\[
\int \frac{\ln x}{x^2}\,dx.
\]

Step 1: Choose a method.

This integrand is a product involving \(\ln x\) and \(x^{-2}\). A very standard approach is integration by parts, because differentiating \(\ln x\) simplifies it, while integrating \(x^{-2}\) also produces a simpler power of \(x\).

Step 2: Set up integration by parts.

Recall the integration by parts formula:

\[
\int u\,dv = uv – \int v\,du.
\]

We choose:

\[
u=\ln x
\]

\[
dv=\frac{1}{x^2}\,dx.
\]

Step 3: Compute \(du\) and \(v\).

Differentiate \(u\):

\[
u=\ln x \quad \Rightarrow \quad du=\frac{1}{x}\,dx.
\]

Integrate \(dv\) to find \(v\):

\[
dv=x^{-2}\,dx \quad \Rightarrow \quad v=\int x^{-2}\,dx.
\]

Now compute \(v\):

\[
\int x^{-2}\,dx=\frac{x^{-1}}{-1}=-x^{-1}=-\frac{1}{x}.
\]

So we have:

\[
v=-\frac{1}{x}.
\]

Step 4: Apply the integration by parts formula.

Substitute into \(\int u\,dv = uv – \int v\,du\):

\[
\int \frac{\ln x}{x^2}\,dx = \left(\ln x\right)\left(-\frac{1}{x}\right) – \int \left(-\frac{1}{x}\right)\left(\frac{1}{x}\,dx\right).
\]

Simplify the first term:

\[
\left(\ln x\right)\left(-\frac{1}{x}\right)=-\frac{\ln x}{x}.
\]

Simplify the integrand inside the remaining integral:

\[
\left(-\frac{1}{x}\right)\left(\frac{1}{x}\,dx\right)=-\frac{1}{x^2}\,dx.
\]

So the formula becomes:

\[
\int \frac{\ln x}{x^2}\,dx = -\frac{\ln x}{x} – \int \left(-\frac{1}{x^2}\right)\,dx.
\]

The minus sign in front and the negative inside cancel:

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

Step 5: Compute the remaining integral.

We already computed \(\int x^{-2}\,dx\) above:

\[
\int \frac{1}{x^2}\,dx=\int x^{-2}\,dx=-\frac{1}{x}.
\]

Step 6: Combine results and add the constant.

Therefore:

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

Factor out \(-\frac{1}{x}\) if desired:

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

Final Answer:

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

See full solution

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Calculus FAQ

What is \(\int \frac{\ln x}{x^2}\,dx\)?

Use integration by parts. Let \(u=\ln x\) and \(dv=x^{-2}\,dx\). Then \(du=\frac{1}{x}\,dx\) and \(v=-\frac{1}{x}\). Now apply the integration by parts formula: \(\int u\,dv=uv-\int v\,du\) So: \(\int \frac{\ln x}{x^2}\,dx=\ln x\left(-\frac{1}{x}\right)-\int\left(-\frac{1}{x}\right)\frac{1}{x}\,dx\) Simplify: \(\int \frac{\ln x}{x^2}\,dx=-\frac{\ln x}{x}+\int\frac{1}{x^2}\,dx\) Since \(\frac{1}{x^2}=x^{-2}\), we have: \(\int x^{-2}\,dx=-x^{-1}=-\frac{1}{x}\) Therefore: \(\int \frac{\ln x}{x^2}\,dx=-\frac{\ln x}{x}-\frac{1}{x}+C\) Factor the numerator: \(\int \frac{\ln x}{x^2}\,dx=-\frac{\ln x+1}{x}+C\) Final answer: \(\int \frac{\ln x}{x^2}\,dx=-\frac{\ln x+1}{x}+C\)

How do you derive \(-\frac{\ln x+1}{x}\)?

Start from integration by parts: \(\int \frac{\ln x}{x^2}\,dx=\ln x\left(-\frac{1}{x}\right)-\int\left(-\frac{1}{x}\right)\frac{1}{x}\,dx\) Simplify the first term: \(\ln x\left(-\frac{1}{x}\right)=-\frac{\ln x}{x}\) Simplify the integral: \(-\int\left(-\frac{1}{x}\right)\frac{1}{x}\,dx=\int\frac{1}{x^2}\,dx\) Now integrate: \(\int\frac{1}{x^2}\,dx=\int x^{-2}\,dx=-x^{-1}=-\frac{1}{x}\) So: \(\int \frac{\ln x}{x^2}\,dx=-\frac{\ln x}{x}-\frac{1}{x}+C\) Combine the terms: \(-\frac{\ln x}{x}-\frac{1}{x}=-\frac{\ln x+1}{x}\) Therefore: \(\int \frac{\ln x}{x^2}\,dx=-\frac{\ln x+1}{x}+C\)

What is \(\int \frac{\ln x}{x^2}\,dx\) for \(x\gt0\)?

For real-valued logarithms, \(\ln x\) requires \(x\gt0\). The antiderivative is still: \(\int \frac{\ln x}{x^2}\,dx=-\frac{\ln x+1}{x}+C\) So, for \(x\gt0\): \(\int \frac{\ln x}{x^2}\,dx=-\frac{\ln x+1}{x}+C\)

Is there a substitution method instead of integration by parts?

Let \(t=\ln x\). Then \(x=e^t\), and \(dx=e^t\,dt\). Rewrite the integrand: \(\frac{\ln x}{x^2}\,dx=\frac{t}{e^{2t}}\cdot e^t\,dt\) Simplify: \(\frac{t}{e^{2t}}\cdot e^t\,dt=te^{-t}\,dt\) So the integral becomes: \(\int te^{-t}\,dt\) This integral still needs integration by parts. It gives the same final result: \(\int \frac{\ln x}{x^2}\,dx=-\frac{\ln x+1}{x}+C\)

What is \(\int \frac{\ln x+1}{x^2}\,dx\)?

Split the fraction into two parts: \(\frac{\ln x+1}{x^2}=\frac{\ln x}{x^2}+\frac{1}{x^2}\) We know: \(\int \frac{\ln x}{x^2}\,dx=-\frac{\ln x+1}{x}+C\) Also: \(\int \frac{1}{x^2}\,dx=-\frac{1}{x}\) So: \(\int \frac{\ln x+1}{x^2}\,dx=-\frac{\ln x+1}{x}-\frac{1}{x}+C\) Simplify: \(\int \frac{\ln x+1}{x^2}\,dx=-\frac{\ln x+2}{x}+C\) Final answer: \(\int \frac{\ln x+1}{x^2}\,dx=-\frac{\ln x+2}{x}+C\)

Can you check by differentiation that \(-\frac{\ln x+1}{x}+C\) is correct?

Let: \(f(x)=-\frac{\ln x+1}{x}\) Rewrite it using a negative exponent: \(f(x)=-(\ln x+1)x^{-1}\) Differentiate using the product rule. \(f'(x)=-\left[\frac{1}{x}\cdot x^{-1}+(\ln x+1)(-1)x^{-2}\right]\) Distribute the negative sign: \(f'(x)=-\frac{1}{x^2}+(\ln x+1)x^{-2}\) Rewrite \(x^{-2}\) as \(\frac{1}{x^2}\). \(f'(x)=-\frac{1}{x^2}+\frac{\ln x+1}{x^2}\) Combine the fractions: \(f'(x)=\frac{-1+\ln x+1}{x^2}\) Simplify the numerator: \(f'(x)=\frac{\ln x}{x^2}\) Since the derivative gives the original integrand, the antiderivative is correct.

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