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∫(x^3 - 7 x) ln xdx
∫ (x^2-1) ln xdx

1 Answer

2 votes

Both integrals can be done by parts.

For the first one:


\displaystyle \int (x^3-7x)\ln(x) \,\mathrm dx = uv - \int v\,\mathrm du

where


u = \ln(x) \implies \mathrm du = \frac{\mathrm dx}x \\\\ \mathrm dv = (x^3-7x)\,\mathrm dx \implies v = \frac{x^4}4-\frac{7x^2}2

Then


\displaystyle \int (x^3-7x)\ln(x)\,\mathrm dx = \left(\frac{x^4}4-\frac{7x^2}2\right)\ln(x) - \int\left(\frac{x^4}4-\frac{7x^2}2\right)\frac{\mathrm dx}x \\\\ \int (x^3-7x)\ln(x)\,\mathrm dx = \left(\frac{x^4}4-\frac{7x^2}2\right)\ln(x) - \int\left(\frac{x^3}4-\frac{7x}2\right)\,\mathrm dx \\\\ \int (x^3-7x)\ln(x)\,\mathrm dx = \boxed{\left(\frac{x^4}4-\frac{7x^2}2\right)\ln(x) - (x^4)/(16) + \frac{7x^2}4 + C}

You can treat the second one identically, and you would end up with


\displaystyle \int(x^2-1)\ln(x)\,\mathrm dx = \left(\frac{x^3}3-x\right)\ln(x) - \int\left(\frac{x^2}3-1\right)\,\mathrm dx \\\\ \int(x^2-1)\ln(x)\,\mathrm dx = \boxed{\left(\frac{x^3}3-x\right)\ln(x) - \frac{x^3}9+\frac{x^2}2} + C}

More generally, the antiderivative of a polynomial
p(x) multiplied by
\ln(x) is given by


\displaystyle \int p(x)\ln(x)\,\mathrm dx = \ln(x)\int p(x)\,\mathrm dx - \int\frac{\displaystyle \int p(u)\,\mathrm du}x\,\mathrm dx

User Kcm
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