Evaluate Using Integration By Parts In Ln X, C X^2 Ln X

evaluateusingintegrationbypartslnx Cx2lnx

Evaluate using integration by parts: ∫ ln x dx.

To compute ∫ ln x dx, we recall the integration by parts formula:

∫ u dv = uv - ∫ v du.

Let us choose u = ln x, which simplifies upon differentiation, and dv = dx, the remaining part of the integral.

Then, du = (1/x) dx, and v = x.

Applying the formula:

∫ ln x dx = x ln x - ∫ x * (1/x) dx = x ln x - ∫ 1 dx = x ln x - x + C.

Therefore, the integral of ln x with respect to x is:

Answer: x ln x - x + C.

Paper For Above instruction

The problem asks for the evaluation of the integral of the natural logarithm function, ln x, using the method of integration by parts. Integration by parts is an essential technique in calculus for integrating the product of two functions, using the formula:

∫ u dv = uv - ∫ v du.

To apply this formula effectively, selecting u and dv is crucial, especially to simplify the integral upon differentiation or integration.

In our specific problem, the integral of ln x is considered, which suggests a strategic choice of u as ln x because its derivative simplifies to 1/x, an easy-to-handle expression. The remaining differential, dx, serves as dv, which integrates straightforwardly to x.

Thus, setting u = ln x and dv = dx gives:

du = (1/x) dx, and v = x.

Applying the integration by parts formula, the integral becomes:

∫ ln x dx = x ln x - ∫ x * (1/x) dx.

The integral ∫ x * (1/x) dx simplifies to ∫ 1 dx, which evaluates to x plus a constant of integration.

Putting it all together yields the final result:

x ln x - x + C.

This demonstrates the powerful utility of integration by parts in addressing integrals involving logarithmic functions, which often simplify through strategic substitutions and choices of u and dv. Mastery of this technique is essential for calculating more complicated integrals in advanced calculus and higher mathematics.

References

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