Mathematics

# Integrate $\displaystyle \int \dfrac{2w+3}{3w+4}dw$.

##### SOLUTION
$\displaystyle I = \int \frac{2w+3}{3w+4}dw = 2\int \frac{w}{3w+4}dw+3\int \frac{dw}{3w+4}$

Let $3w+4 = u \Rightarrow du = 3dw$

$\displaystyle I = 2\int \frac{(u-4)}{3u} \frac{du}{3}+3\int \dfrac 1 {3u} du+c$

$\displaystyle = \frac{2}{9} \left( u-4 log u \right )+\frac{3log \left | 3w+4 \right |}{3}+c$

$\displaystyle = \frac{2}{9}(3w+4)-\frac{-8}{9}log\left | 3w+4 \right |+\frac{3 log\left | 3w+4 \right |}{3}+c$

$\displaystyle I = \dfrac{2}{9}(3w+4)+\frac{1}{9}log \left | 3w+4 \right |+c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
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If $\displaystyle \int \left[\left(\dfrac{x}{e}\right)^x + \left(\dfrac{e}{x} \right)^x \right] \ln x \, dx = A \left(\dfrac{x}{e}\right)^x + B \left(\dfrac{e}{x} \right)^x + C$, then value of $A + B =$?