Evaluate the following integrals: (4x+1) x^2-x-2 dx - Vidyadip

Question

Evaluate the following integrals:
$\int(4\text{x}+1)\sqrt{\text{x}^2-\text{x}-2}\text{dx}$

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Answer

Let $\text{I}=\int(4\text{x}+1)\sqrt{\text{x}^2-\text{x}-2}\text{dx}$
Let $4\text{x}+1=\lambda\frac{\text{d}}{\text{dx}}(\text{x}^2-\text{x}-2)+\mu$
$=\lambda(2\text{x}-1)+\mu$
Equating similar terms, we get,
$2\lambda=4\Rightarrow\lambda=2$
$-\lambda+\mu=1\Rightarrow\mu=3$
So,
$\text{I}=\int(2(2\text{x}-1)+3)\sqrt{\text{x}^2-\text{x}-2}\text{dx}$
$=2\int(2\text{x}-1)\sqrt{\text{x}^2-\text{x}-2}\text{dx}+3\int\sqrt{\text{x}^2-\text{x}-2}\text{dx}$
Let $\text{x}^2-\text{x}-2=\text{t}$
$(2\text{x}-1)\text{dx = dt}$
$\therefore\ \text{I}=2\int\sqrt{\text{t}}\text{dt}+3\int\sqrt{\Big(\text{x}-\frac{1}{2}\Big)^2-\Big(\frac{3}{2}\Big)^2}\text{dx}$
$\Rightarrow\text{I}=2\frac{\text{t}^{\frac{3}{2}}}{\frac{3}{2}}+3\begin{Bmatrix}\frac{\Big(\text{x}-\frac{1}{2}\Big)}{2}\sqrt{\text{x}^2-\text{x}-2}\\-\frac{9}{8}\log\Big|\Big(\text{x}-\frac{1}{2}\Big)+\sqrt{\text{x}^2-\text{x}-2}\Big|\end{Bmatrix}+\text{C}$
Hence,
$\Rightarrow\text{I}=\frac{4}{3}(\text{x}^2-\text{x}-2)^{\frac{3}{2}}+\frac{3}{4}(2\text{x}-1)\sqrt{\text{x}^2-\text{x}-2}\\-\frac{27}{8}\log\Big|\Big(\text{x}-\frac{1}{2}\Big)+\sqrt{\text{x}^2-\text{x}-2}\Big|+\text{C}$

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