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Algebra of Derivatives

Evaluate ∫4...

Question

Evaluate $\int \frac{4 x + 3}{2 x^{2} + 2 x + 5} d x$ .

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Solution

We need to evaluate $\int \frac{4 x + 3}{2 x^{2} + 2 x + 5} d x$

Consider $\int \frac{4 x + 3}{2 x^{2} + 2 x + 5} d x = \int \frac{4 x}{2 x^{2} + 2 x + 5} d x + \int \frac{3}{2 x^{2} + 2 x + 5} d x . . . (1)$

Consider $\int \frac{4}{2 x^{2} + 2 x + 5} d x = 4 \int \frac{1}{2 x^{2} + 2 x + 5} d x$

$= 4 \int \frac{x}{{(\sqrt{2} x + \frac{1}{\sqrt{2}})}^{2} + \frac{9}{2}} d x$

Let $u = \sqrt{2} x + \frac{1}{\sqrt{2}} \Rightarrow d u = \sqrt{2} d x$

$= 4 \int \frac{\sqrt{2} u - 1}{\sqrt{2} (2 u^{2} + 9)} d u$

$= 4 \cdot \frac{1}{\sqrt{2}} \int \frac{\sqrt{2} u - 1}{2 u^{2} + 9} d u$

$= \frac{4}{\sqrt{2}} (\int \frac{\sqrt{2} u}{2 u^{2} + 9} d u - \int \frac{1}{2 u^{2} + 9} d u)$

$= \frac{4}{\sqrt{2}} (\frac{1}{2 \sqrt{2}} ln | 2 u^{2} + 9 | + \frac{1}{3 \sqrt{2}} {tan}^{- 1} (\frac{\sqrt{2}}{3} u))$

Substitute back $u = \sqrt{2} x + \frac{1}{\sqrt{2}}$ we get

$= \frac{4}{\sqrt{2}} ⎛ ⎝ \frac{1}{2 \sqrt{2}} ln ∣ ∣ ∣ ∣ 2 {(\sqrt{2} x + \frac{1}{\sqrt{2}})}^{2} + 9 ∣ ∣ ∣ ∣ + \frac{1}{3 \sqrt{2}} {tan}^{- 1} (\frac{\sqrt{2}}{3} \cdot (\sqrt{2} x + \frac{1}{\sqrt{2}})) ⎞ ⎠$

$∴ \int \frac{4}{2 x^{2} + 2 x + 5} d x = ln | 4 x^{2} + 4 x + 10 | - \frac{2}{3} {tan}^{- 1} (\frac{2}{3} x + \frac{1}{3}) . . . (2)$

Now consider $\int \frac{3}{2 x^{2} + 2 x + 5} d x = 3 \int \frac{1}{2 x^{2} + 2 x + 5} d x$

$= 3 \int \frac{1}{{(\sqrt{2} x + \frac{1}{\sqrt{2}})}^{2} + \frac{9}{2}} d x$

Let $u = \sqrt{2} x + \frac{1}{\sqrt{2}} \Rightarrow d u = \sqrt{2} d x$

$= 3 \int \frac{\sqrt{2}}{2 u^{2} + 9} d u$

$= 3 \sqrt{2} \int \frac{1}{2 u^{2} + 9} d u$

$= 3 \sqrt{2} \frac{1}{3 \sqrt{2}} {tan}^{- 1} (\frac{\sqrt{2}}{3} u)$

Substitute back $u = \sqrt{2} x + \frac{1}{\sqrt{2}}$ we get

$= {tan}^{- 1} (\frac{\sqrt{2}}{3} \cdot (\sqrt{2} x + \frac{1}{\sqrt{2}}))$

$∴ \int \frac{3}{2 x^{2} + 2 x + 5} d x = 3 \int \frac{1}{2 x^{2} + 2 x + 5} d x = {tan}^{- 1} (\frac{2}{3} x + \frac{1}{3}) . . . (3)$

Substitute $(2), (3)$ in $(1)$ we get

$\int \frac{4 x + 3}{2 x^{2} + 2 x + 5} d x = ln | 4 x^{2} + 4 x + 10 | - \frac{2}{3} {tan}^{- 1} (\frac{2}{3} x + \frac{1}{3}) + {tan}^{- 1} (\frac{2}{3} x + \frac{1}{3})$

$∴ \int \frac{4 x + 3}{2 x^{2} + 2 x + 5} d x = ln | 4 x^{2} + 4 x + 10 | + \frac{1}{3} {tan}^{- 1} (\frac{2}{3} x + \frac{1}{3}) + c$ where $c$ is constant of integration.

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