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Question Number 185837 by test1234 last updated on 28/Jan/23

$$\int \frac{cx+b^2}{2x}-\frac{cx+2b^2}{4x}+\frac{{\left(cx\right)}^2+4b^2}{8x}dx=?$$

Commented by MJS_new last updated on 28/Jan/23

$$\mathrm{seriously}?$$$$=\frac{c^2}{8}\int xdx+\frac{c}{4}\int dx+\frac{b^2}{2}\int \frac{dx}{x}=$$$$=\frac{c^2}{16}{x}^2+\frac{c}{4}x+\frac{b^2}{2}\ln \mid x\mid +C$$

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