Show that -2 x-1-2 x-3 d x-x-log-2 x-32-C

∫2x 1/2x+3 dx = x log|(2x + 3)^2+ C Let I = ∫2x 1/2x+3 dx =∫2x + 3 3 1/2x + 3 dx = ∫ 1dx 4∫1/2x + 3dx = x ∫4/2(x+3/2 )dx =x 2 log+|(x+3/2 )|C' = x ...

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Byju's Answer

Standard XII

Mathematics

First Fundamental Theorem of Calculus

Show that ∫2 ...

Question

Show that $\int \frac{2 x - 1}{2 x + 3} d x = x - l o g | (2 x + 3)^{2} |$ + C

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Solution

$\int \frac{2 x - 1}{2 x + 3} d x = x - l o g | (2 x + 3)^{2} + C L e t I = \int \frac{2 x - 1}{2 x + 3} d x = \int \frac{2 x + 3 - 3 - 1}{2 x + 3} d x = \int 1 d x - 4 \int \frac{1}{2 x + 3} d x = x - \int \frac{4}{2 (x + \frac{3}{2})} d x = x - 2 l o g + | (x + \frac{3}{2}) | C^{'} = x - 2 l o g (\frac{2 x + 3}{2}) | + C^{'} x - 2 l o g (2 x + 3) | + 2 l o g 2 + C^{'} [∵ l o g \frac{m}{n} = l o g m - l o g n] = x - l o g | (2 x + 3)^{2} | + C$

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Show that ∫2x−12x+3dx=x−log|(2x+3)2| + C

∫2x−12x+3dx=x−log|(2x+3)2+CLetI=∫2x−12x+3dx=∫2x+3−3−12x+3dx=∫1dx−4∫12x+3dx=x−∫42(x+32)dx=x−2 log+|(x+32)|C′=x−2 log(2x+32)|+C′x−2 log(2x+3)|+2 log2+C′[∵logmn=log m−log n]=x−log|(2x+3)2|+C

Show that $\int \frac{2 x - 1}{2 x + 3} d x = x - l o g | (2 x + 3)^{2} |$ + C