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

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Answer 1 · 2016-12-28T14:39:04

By the laws of logarithms:

$=intlnx^2 + lnsqrt(4x - 1)dx$

$=intlnx^2dx + intlnsqrt(4x - 1)dx$

$=2intlnxdx+ 1/2intln(4x - 1)dx$

We let $u = 4x - 1$ . Then $du = 4dx$ and $dx = (du)/4$ .

$=2intlnxdx + 1/2intlnu (du)/4$

$=2intlnxdx + 1/8intlnudu$

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

$int(udv) = uv - int(vdu)$

$int(lnx) = xlnx - int(x * 1/x)dx$

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

Back to the original integral.

$=2(xlnx - x) + 1/8(u lnu -u)+ C$

$=2xlnx - 2x + 1/8(4x - 1ln(4x - 1) - (4x - 1)) + C$

$=2xlnx - 2x + ((4x - 1)ln(4x - 1) - 4x + 1)/8 + C$

Hopefully this helps!

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