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It's based on integrals....​

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[tex] \huge \boxed{\mathbb{QUESTION} \downarrow}[/tex]

[tex] \large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}[/tex]

[tex] \tt \:\int \: \frac{ x + 2 }{( {x}^{2} + 3x + 3) \sqrt{x + 1}} \: dx \\ [/tex]

First, let's take I as ⇨ [tex]\tt \:\int \: \frac{ x + 2 }{( {x}^{2} + 3x + 3) \sqrt{x + 1}} \: dx \\ [/tex].

[tex]\tt \:I = \int \: \frac{ x + 2 }{( {x}^{2} + 3x + 3) \sqrt{x +1}} \: dx \\\tt \: I =\int \: \frac{ x + 2 }{( {x}^{2} + 2x + 1 + x + 2) \sqrt{x + 1}} \: dx \\ \tt \:I =\int \: \frac{ x + 2 }{( ({x + 1}^{2}) + x + 2) \sqrt{x + 1}}[/tex]

Let, x + 1 = m² => dx = 2mdm.

[tex]\tt \:I =\int \: \frac{ {m}^{2} + 1}{ {m}^{4} + {m}^{2} + 1 \cdot \: m} 2mdm \\ \tt \:I =\int \: \frac{ {m}^{2} + 1}{ {m}^{4} + {m}^{2} + 1 \cdot \: \bcancel{ m}} 2 \bcancel{m}dm \\ \tt \: I = \: \int \: \frac{ {m}^{2} + 1}{ {m}^{4} + {m}^{2} + 1 } 2dm[/tex]

Now, divide the numerator & denominator by m²....we'll get it as...

[tex]\tt \:I =2\int \: \frac{ 1 + \frac{1}{ {m}^{2} } }{ {m}^{2} + 1 + \frac{1}{ {m}^{2} } } \: dm \\\tt \: I =2\int \: \frac{ 1 + \frac{1}{ {m}^{2} } }{( {m}^{2} + \frac{1}{ {m}^{2} } - 2) + 3} \: dm \\\tt \: I =2\int \: \frac{ (1 + \frac{1}{ {m}^{2} }) \: dm }{ ({m} - \frac{1}{ m } ) ^{2} + 3} [/tex]

Now, let m - 1/m be t => (1 + 1/m²) dm = dt

[tex]\tt \:I =2\int \: \frac{ dt}{ {t}^{2} + 3 } \\ \tt \:I = 2\int \: \frac{ dt}{ {t}^{2} + ( \sqrt{3}) ^{2} }[/tex]

We know, [tex]\tt \:\int \: \frac{dx}{ {x}^{2} + a ^{2} } = \frac{1}{a} tan ^{ - 1} (\frac{x}{a} ) + c[/tex]...therefore...

[tex]\tt \:I = \frac{2}{ \sqrt{3} } {tan}^{ - 1} ( \frac{t}{ \sqrt{3} } ) + c \: \rightarrow \boxed{ \tt \: eq. \: 1}[/tex]

Now, substitute the value of 't' in eq. 1..we'll get..

[tex]\tt \:I = \frac{2}{ \sqrt{3} } {tan}^{ - 1} ( \frac{m - \frac{1}{m} }{ \sqrt{3} } ) + c \: \rightarrow \boxed{ \tt \: eq. \: 2}[/tex]

Now, substitute the value of 'm' in eq. 2...we'll get...

[tex] \tt \: I = \frac{2}{ \sqrt{3} } {tan}^{ - 1} ( \frac{ \sqrt{x + 1} - \frac{1}{ \sqrt{x - 1} } }{ \sqrt{3} } ) + c \\ \tt \:I = \frac{2}{ \sqrt{3} } {tan}^{ - 1} ( \frac{x + 1 - 1}{ \sqrt{3} \sqrt{x - 1} } ) + c \\ \boxed{\boxed{ \bf \: I = \frac{2}{ \sqrt{3} } {tan}^{ - 1} ( \frac{x }{ \sqrt{3 (x - 1} )} ) + c }}[/tex]

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