The integral ∫(2x+5)(x+1)(x+2)(x+3)(x+4)dx is equal to

I=∫(2x+5)(x+1)(x+2)(x+3)(x+4)dx put x2+5x+5=t⇒2x+5dx=dt I=∫(2x+5)x2+5x+4x2+5x+6dx =∫dtt-1t+1=12logt-1t+1+c =12logx2+5x+4x2+5x+6+c =12ln⁡(x+1)(x+4)(x+2)(x+3)+c

The integral ∫(2x+5)(x+1)(x+2)(x+3)(x+4)dx is equal to

I=∫(2x+5)(x+1)(x+2)(x+3)(x+4)dx put x2+5x+5=t⇒2x+5dx=dt I=∫(2x+5)x2+5x+4x2+5x+6dx =∫dtt-1t+1=12logt-1t+1+c =12logx2+5x+4x2+5x+6+c =12ln⁡(x+1)(x+4)(x+2)(x+3)+c

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The integral $\int \frac{(2 x + 5)}{(x + 1) (x + 2) (x + 3) (x + 4)} dx$ is equal to

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$\frac{1}{2} \ln |\frac{3 x + 4}{5 x + 6}| + c$

$\frac{1}{2} \ln |\frac{x^{2} + 3 x + 4}{x^{2} + 5 x + 6}| + c$

$\frac{1}{2} \ln |\frac{x^{2} + 5 x + 4}{x^{2} + 5 x + 6}| + c$

$\frac{1}{2} \ln |\frac{(x + 1) (x + 4)}{(x + 2) (x + 3)}| + c$

answer is C, D.

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Detailed Solution

$I = \int \frac{(2 x + 5)}{(x + 1) (x + 2) (x + 3) (x + 4)} dx p u t x^{2} + 5 x + 5 = t \Rightarrow (2 x + 5) dx = dt I = \int \frac{(2 x + 5)}{(x^{2} + 5 x + 4) (x^{2} + 5 x + 6)} dx = \int \frac{dt}{(t - 1) (t + 1)} = \frac{1}{2} \log |\frac{t - 1}{t + 1}| + c = \frac{1}{2} \log |\frac{x^{2} + 5 x + 4}{x^{2} + 5 x + 6}| + c = \frac{1}{2} \ln |\frac{(x + 1) (x + 4)}{(x + 2) (x + 3)}| + c$