Indefinite Integrals — Maths STD 12 Science — Question

We express $\text{x}+3=\text{A}\frac{\text{d}}{\text{dx}}(3-4\text{x}-\text{x}^2)+\text{B}$

$\text{x}+3=\text{A}(-4-2\text{x})+\text{B}$

Equating the co-efficient of x and constants, we get

$1=-2\text{A}$ and $3=-4\text{A + B}$

or $\text{A}=-\frac{1}{2}$ and $\text{B}=1$

$\therefore\ \text{I}=\int\Big(-\frac{1}{2}(-4-2\text{x})+1\Big)\sqrt{3-4\text{x}-\text{x}^2}\text{dx}$

$=-\frac{1}{2}\int(-4-2\text{x})\sqrt{3-4\text{x}-\text{x}^2}\text{dx}+\int\sqrt{3-4\text{x}-\text{x}^2}\text{dx}$

$=-\frac{1}{2}\text{I}_1+\text{I}_2\ \dots(1)$

Now, $\text{I}_1=\int(-4-2\text{x})\sqrt{3-4\text{x}-\text{x}^2}\text{dx}$

Let $3-4\text{x}-\text{x}^2=\text{u}$

On differentiating both sides, we get

$(-4-2\text{x})\text{dx = du}$

$\therefore\ \text{I}_1=\int\sqrt{\text{u}}\text{du}$

$=\frac{2}{3}\text{u}^{\frac{3}{2}}+\text{c}_1$

$=\frac{2}{3}(3-4\text{x}-\text{x}^2)^{\frac{3}{2}}+\text{c}_1\ \dots(2)$

And, $\text{I}_2=\int\sqrt{3-4\text{x}-\text{x}^2}\text{dx}$

$=\int\sqrt{3+4-4-4\text{x}-\text{x}^2}\text{dx}$

$=\int\sqrt{(\sqrt7)^2-(\text{x}+2)^2}\text{dx}$

Let $(\text{x}+2)=\text{u}$

On differentiating both sides, we get

$\text{dx = du}$

$\therefore\ \text{I}_2=\int\sqrt{(\sqrt7)^2-(\text{u})^2}\text{du}$

$=\frac{\text{u}}{2}\sqrt{(\sqrt7)^2-(\text{u})^2}+\frac{1}{2}(\sqrt7)^2\sin^{-1}\Big(\frac{\text{u}}{\sqrt7}\Big)+\text{c}_2$

$=\frac{\text{x}+2}{2}\sqrt{7-(\text{x}+2)^2}+\frac{7}{2}\sin^{-1}\Big(\frac{\text{x}+2}{\sqrt7}\Big)+\text{c}_2$

$=\frac{\text{x}+2}{2}\sqrt{3-4\text{x}-\text{x}^2}+\frac{7}{2}\sin^{-1}\Big(\frac{\text{x}+2}{\sqrt7}\Big)+\text{c}_2\ \dots(3)$

From (1), (2) and (3), we get

$\therefore\ \text{I}=-\frac{1}{2}\Big(\frac{2}{3}(3-4\text{x}-\text{x}^2)^{\frac{3}{2}}+\text{c}_1\Big)\\+\Big(\frac{\text{x}+2}{2}\sqrt{3-4\text{x}-\text{x}^2}+\frac{7}{2}\sin^{-1}\Big(\frac{\text{x}+2}{\sqrt7}\Big)+\text{c}_2\Big)$

$=-\frac{1}{3}(3-4\text{x}-\text{x}^2)^{\frac{3}{2}}+\frac{\text{x}+2}{2}\sqrt{3-4\text{x}-\text{x}^2}\\+\frac{7}{2}\sin^{-1}\Big(\frac{\text{x}+2}{\sqrt7}\Big)+\text{C}$

Hence,

$\int(\text{x}+3)\sqrt{3-4\text{x}-\text{x}^2}\text{dx}=-\frac{1}{3}(3-4\text{x}-\text{x}^2)^{\frac{3}{2}}\\+\frac{\text{x}+2}{2}\sqrt{3-4\text{x}-\text{x}^2}+\frac{7}{2}\sin^{-1}\Big(\frac{\text{x}+2}{\sqrt7}\Big)+\text{C}$