Question

Evaluate the following:

$\int (x + 1) \sqrt{2 x^{2} + 3} d x$ .

Accepted Answer

Let $\int (x + 1) \sqrt{2 x^{2} + 3} d x$

Let $x + 1 = A \frac{d}{d x} (2 x^{2} + 3) + B$

$\Rightarrow x + 1 = A (4 x) + B$

$\Rightarrow x + 1 = 4 A x + B$

Comparing coefficient of $x$ and constant term from both sides, then

$4 A = 1 and B = 1$

$\Rightarrow A = \frac{1}{4} and B = 1$

Therefore, $I = \int [\frac{1}{4} (4 x) + 1] (\sqrt{2 x^{2} + 3}) d x$

$\Rightarrow I = \frac{1}{4} \int 4 x (\sqrt{2 x^{2} + 3}) d x + \int (\sqrt{2 x^{2} + 3}) d x$

$\Rightarrow I = I_{1} + I_{2}$

Now, consider $I_{1} = \frac{1}{4} \int 4 x (\sqrt{2 x^{2} + 3}) d x$

Using the formula: $\int {[f (x)]}^{n} f^{'} (x) d x = \frac{{[f (x)]}^{n + 1}}{n + 1} + c$

Therefore, $I_{1} = \frac{1}{4} \frac{{(2 x^{2} + 3)}^{\frac{3}{2}}}{\frac{3}{2}} + C_{1}$

$\Rightarrow I_{1} = \frac{1}{6} {(2 x^{2} + 3)}^{\frac{3}{2}} + C_{1}$

Similarly, consider, $I_{2} = \int (\sqrt{2 x^{2} + 3}) d x = \sqrt{2} \int \sqrt{x^{2} + {(\sqrt{\frac{3}{2}})}^{2}} d x$

Using the formula: $\int (\sqrt{x^{2} + a^{2}}) d x = \frac{x}{2} \sqrt{x^{2} + a^{2}} + \frac{a^{2}}{2} \log |x + \sqrt{x^{2} + a^{2}}| + c$

Therefore, $I_{2} = \sqrt{2} [\frac{x}{2} \sqrt{x^{2} + {(\sqrt{\frac{3}{2}})}^{2}} + \frac{3}{4} \log |x + \sqrt{x^{2} + {(\sqrt{\frac{3}{2}})}^{2}}|] + C_{2}$

$\Rightarrow I_{2} = \frac{x}{2} \sqrt{2 x^{2} + 3} + \frac{3}{2 \sqrt{2}} \log |x + \sqrt{x^{2} + \frac{3}{2}}| + C_{2}$

Hence,

$I = \frac{1}{6} {(2 x^{2} + 3)}^{\frac{3}{2}} + \frac{x}{2} \sqrt{2 x^{2} + 3} + \frac{3}{2 \sqrt{2}} \log |x + \sqrt{x^{2} + \frac{3}{2}}| + C$ , where $C_{1} + C_{2} = C$ .

R D Sharma Solutions for Chapter: Indefinite Integrals, Exercise 1: EXERCISE

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