Evaluate the integral. int-4x^(-2)ln(x)dx Answer:

Write Integral: Let's denote the integral by I and start by writing it down: I = ∫-4x^(-2)ln(x)dx We can simplify the integral by factoring out the constant -4: I = -4∫x^(-2)ln(x)dx Now we have an integral of the form ∫x^n ln(x)dx, which can be solved using integration by parts. Integration by parts formula is ∫u dv = uv - ∫v du, where we need to choose u and dv. Let's choose u = ln(x) and dv = x^(-2)dx. Then we need to compute du and v: du = (1/x)dx and v = ∫x^(-2)dx = -x^(-1) Now we can apply the integration by parts formula.Simplify Integral: Applying integration by parts: I = -4(uv - ∫v du) I = -4(ln(x)(-x^(-1)) - ∫(-x^(-1))(1/x)dx) Simplify the expression: I = 4x^(-1)ln(x) - ∫(-1/x^2)dx The integral of -1/x^2 is 1/x, so we can integrate that directly.Apply Integration by Parts: Continuing with the integration: I = 4x^(-1)ln(x) + 4∫(1/x^2)dx I = 4x^(-1)ln(x) + 4(-1/x) Now we can simplify the expression: I = 4/x * ln(x) - 4/x + C, where C is the constant of integration. This is the final answer for the indefinite integral.

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Evaluate the integral.

int-4x^(-2)ln(x)dx
Answer:

Evaluate the integral. $\newline$ $\int-4 x^{-2} \ln (x) d x$ $\newline$ Answer:

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Calculus

Find indefinite integrals using the substitution and by parts

Full solution

Q. Evaluate the integral.

\newline

\int-4 x^{-2} \ln (x) d x

\newline

Answer:

Write Integral: Let's denote the integral by $I$ and start by writing it down: $\newline$ $I = \int -4x^{-2}\ln(x)\,dx$ $\newline$ We can simplify the integral by factoring out the constant $-4$ : $\newline$ $I = -4\int x^{-2}\ln(x)\,dx$ $\newline$ Now we have an integral of the form $\int x^n \ln(x)\,dx$ , which can be solved using integration by parts. $\newline$ Integration by parts formula is $\int u \, dv = uv - \int v \, du$ , where we need to choose $u$ and $dv$ . $\newline$ Let's choose $u = \ln(x)$ and $dv = x^{-2}\,dx$ . $\newline$ Then we need to compute $I = \int -4x^{-2}\ln(x)\,dx$ $0$ and $I = \int -4x^{-2}\ln(x)\,dx$ $1$ : $\newline$ $I = \int -4x^{-2}\ln(x)\,dx$ $2$ and $I = \int -4x^{-2}\ln(x)\,dx$ $3$ $\newline$ Now we can apply the integration by parts formula.
Simplify Integral: Applying integration by parts: $\newline$ $I = -4(uv - \int v du)$ $\newline$ $I = -4(\ln(x)(-x^{-1}) - \int(-x^{-1})(1/x)dx)$ $\newline$ Simplify the expression: $\newline$ $I = 4x^{-1}\ln(x) - \int(-1/x^2)dx$ $\newline$ The integral of $-1/x^2$ is $1/x$ , so we can integrate that directly.
Apply Integration by Parts: Continuing with the integration: $\newline$ $I = 4x^{-1}\ln(x) + 4\int(\frac{1}{x^2})dx$ $\newline$ $I = 4x^{-1}\ln(x) + 4(-\frac{1}{x})$ $\newline$ Now we can simplify the expression: $\newline$ $I = \frac{4}{x} \cdot \ln(x) - \frac{4}{x} + C$ , where $C$ is the constant of integration. $\newline$ This is the final answer for the indefinite integral.