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Indefinite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
Evaluate the following integrals:
$\int(2\text{x}+3)\sqrt{\text{x}^2+4\text{x}+3}\text{dx}$
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Answer

Let $\text{I}=\int(2\text{x}+3)\sqrt{\text{x}^2+4\text{x}+3}\text{dx}$Let $(2\text{x}+3)=\lambda\frac{\text{d}}{\text{dx}}(\text{x}^2+4\text{x}+3)+\mu$
$=\lambda(2\text{x}+4)+\mu$ Equating similar terms, we get, $\lambda=1\text{ and }4\lambda+\mu=3$ $\Rightarrow\mu=-1$ So, $\text{I}=\int((2\text{x}+4)+(-1))\sqrt{\text{x}^2+4\text{x}+3}\text{dx}$ $=\int(2\text{x}+4)\sqrt{\text{x}^2+4\text{x}+3}\text{dx}-\int\sqrt{\text{x}^2+4\text{x}+3}\text{dx}$ Let $\text{x}^2+4\text{x}+3=\text{t}$ $\Rightarrow(2\text{x}+4)\text{dx = dt}$ $\therefore\ \text{I}=\int\sqrt{\text{t}}\text{dt}-\int\sqrt{(\text{x}+2)^2-1}\text{dx}$ $=\frac{3}{2}\text{t}^{\frac{3}{2}}-\frac{(\text{x}+2)}{2}\sqrt{\text{x}^2+4\text{x}+3}\\+\frac{1}{2}\log\Big|(\text{x}+2)+\sqrt{\text{x}^2+4\text{x}+3}\Big|+\text{C}$ Hence, $\text{I}=\frac{2}{3}(\text{x}^2+4\text{x}+3)^{\frac{3}{2}}-\Big(\frac{\text{x}+2}{2}\Big)\sqrt{\text{x}^2+4\text{x}+3}\\+\frac{1}{2}\log\Big|(\text{x}+2)+\sqrt{\text{x}^2+4\text{x}+3}\Big|+\text{C}$

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