How do you evaluate the integral $\int ln (x^{2} \sqrt{4 x - 1})$ $int ln(x^2sqrt(4x-1))$ ?

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How do you evaluate the integral $\int ln (x^{2} \sqrt{4 x - 1})$ $int ln(x^2sqrt(4x-1))$ ?

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Answer 1 · 2016-12-28T14:39:04

By the laws of logarithms:

$= \int {ln x}^{2} + ln \sqrt{4 x - 1} d x$ $=intlnx^2 + lnsqrt(4x - 1)dx$

$= \int {ln x}^{2} d x + \int ln \sqrt{4 x - 1} d x$ $=intlnx^2dx + intlnsqrt(4x - 1)dx$

$= 2 \int ln x d x + \frac{1}{2} \int ln (4 x - 1) d x$ $=2intlnxdx+ 1/2intln(4x - 1)dx$

We let $u = 4 x - 1$ $u = 4x - 1$ . Then $d u = 4 d x$ $du = 4dx$ and $d x = \frac{d u}{4}$ $dx = (du)/4$ .

$= 2 \int ln x d x + \frac{1}{2} \int ln u \frac{d u}{4}$ $=2intlnxdx + 1/2intlnu (du)/4$

$= 2 \int ln x d x + \frac{1}{8} \int ln u d u$ $=2intlnxdx + 1/8intlnudu$

We need to integrate the natural logarithm function by parts. Let $u = ln x$ $u = lnx$ and $d v = 1 d x$ $dv = 1dx$ . Then $d u = \frac{1}{x} d x$ $du = 1/xdx$ and $v = x$ $v = x$ .

$\int (u d v) = u v - \int (v d u)$ $int(udv) = uv - int(vdu)$

$\int (ln x) = x ln x - \int (x \cdot \frac{1}{x}) d x$ $int(lnx) = xlnx - int(x * 1/x)dx$

$\int (ln x) = x ln x - x + C$ $int(lnx) = xlnx - x + C$

Back to the original integral.

$= 2 (x ln x - x) + \frac{1}{8} (u ln u - u) + C$ $=2(xlnx - x) + 1/8(u lnu -u)+ C$

$= 2 x ln x - 2 x + \frac{1}{8} (4 x - 1 ln (4 x - 1) - (4 x - 1)) + C$ $=2xlnx - 2x + 1/8(4x - 1ln(4x - 1) - (4x - 1)) + C$

$= 2 x ln x - 2 x + \frac{(4 x - 1) ln (4 x - 1) - 4 x + 1}{8} + C$ $=2xlnx - 2x + ((4x - 1)ln(4x - 1) - 4x + 1)/8 + C$

Hopefully this helps!

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How do you evaluate the integral $\int ln (x^{2} \sqrt{4 x - 1})$ $int ln(x^2sqrt(4x-1))$ ?

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