Question

Question :-

$find \: \int \sqrt{ {x}^{2} + 2x + 5 \: dx}$

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Answer 1

We're asked to evaluate,

$\displaystyle\longrightarrow I=\int\sqrt{x^2+2x+5}\ dx$

Taking $5=1+4,$

$\displaystyle\longrightarrow I=\int\sqrt{x^2+2x+1+4}\ dx$

$\displaystyle\longrightarrow I=\int\sqrt{(x+1)^2+2^2}\ dx$

$\displaystyle\longrightarrow I=\int\sqrt{2^2\left[\dfrac{(x+1)^2}{2^2}+1\right]}\ dx$

$\displaystyle\longrightarrow I=2\int\sqrt{\left(\dfrac{x+1}{2}\right)^2+1\right]}\ dx\quad\quad\dots(1)$

Put,

$\longrightarrow\dfrac{x+1}{2}=\tan\theta$

$\longrightarrow x=2\tan\theta-1$

$\longrightarrow dx=2\sec^2\theta\ d\theta$

Also,

$\longrightarrow\sec\theta=\sqrt{\tan^2\theta+1}$

$\longrightarrow\sec\theta=\sqrt{\left(\dfrac{x+1}{2}\right)^2+1}$

$\longrightarrow\sec\theta=\dfrac{1}{2}\sqrt{x^2+2x+5}$

Then (1) becomes,

$\displaystyle\longrightarrow I=4\int\sqrt{\tan^2\theta+1\right]}\ \sec^2\theta\ d\theta$

$\displaystyle\longrightarrow I=4\int\sec^3\theta\ d\theta\quad\quad\dots(1)$

$\displaystyle\longrightarrow I=4\int\sec\theta\ \sec^2\theta\ d\theta$

Performing integration by parts,

$\displaystyle\longrightarrow I=4\left[\sec\theta\tan\theta-\int \sec\theta\tan^2\theta\ d\theta\right]$

$\displaystyle\longrightarrow I=4\sec\theta\tan\theta-4\int \sec\theta(\sec^2\theta-1)\ d\theta$

$\displaystyle\longrightarrow I=4\sec\theta\tan\theta-4\int(\sec^3\theta-\sec\theta)\ d\theta$

$\displaystyle\longrightarrow I=4\sec\theta\tan\theta-4\int\sec^3\theta\ d\theta+4\int\sec\theta\ d\theta\quad\quad\dots(2)$

Adding (1) and (2) we get,

$\displaystyle\longrightarrow 2I=4\sec\theta\tan\theta+4\int\sec\theta\ d\theta$

$\displaystyle\longrightarrow I=2\sec\theta\tan\theta+2\log|\sec\theta+\tan\theta|+C'$

Undoing substitutions for $\sec\theta$ and $\tan\theta,$

$\displaystyle\longrightarrow I=\dfrac{1}{2}(x+1)\sqrt{x^2+2x+5}+2\log\left|\dfrac{x+1+\sqrt{x^2+2x+5}}{2}\right|+C'$

Taking $C'-2\log2=C,$

$\displaystyle\longrightarrow\underline{\underline{I=\dfrac{1}{2}(x+1)\sqrt{x^2+2x+5}+2\log\left|x+1+\sqrt{x^2+2x+5}\right|+C}}$

Answer 2

Answer:

We're asked to evaluate,

\displaystyle\longrightarrow I=\int\sqrt{x^2+2x+5}\ dx⟶I=∫

x

2

+2x+5

dx

Taking 5=1+4,5=1+4,

\displaystyle\longrightarrow I=\int\sqrt{x^2+2x+1+4}\ dx⟶I=∫

x

2

+2x+1+4

dx

\displaystyle\longrightarrow I=\int\sqrt{(x+1)^2+2^2}\ dx⟶I=∫

(x+1)

2

+2

2

dx

\displaystyle\longrightarrow I=\int\sqrt{2^2\left[\dfrac{(x+1)^2}{2^2}+1\right]}\ dx⟶I=∫

2

2

[

2

2

(x+1)

2

+1]

dx

\displaystyle\longrightarrow I=2\int\sqrt{\left(\dfrac{x+1}{2}\right)^2+1\right]}\ dx\quad\quad\dots(1)

Put,

\longrightarrow\dfrac{x+1}{2}=\tan\theta⟶

2

x+1

=tanθ

\longrightarrow x=2\tan\theta-1⟶x=2tanθ−1

\longrightarrow dx=2\sec^2\theta\ d\theta⟶dx=2sec

2

θ dθ

Also,

\longrightarrow\sec\theta=\sqrt{\tan^2\theta+1}⟶secθ=

tan

2

θ+1

\longrightarrow\sec\theta=\sqrt{\left(\dfrac{x+1}{2}\right)^2+1}⟶secθ=

(

2

x+1

)

2

+1

\longrightarrow\sec\theta=\dfrac{1}{2}\sqrt{x^2+2x+5}⟶secθ=

2

1

x

2

+2x+5

Then (1) becomes,

\displaystyle\longrightarrow I=4\int\sqrt{\tan^2\theta+1\right]}\ \sec^2\theta\ d\theta

\displaystyle\longrightarrow I=4\int\sec^3\theta\ d\theta\quad\quad\dots(1)⟶I=4∫sec

3

θ dθ…(1)

\displaystyle\longrightarrow I=4\int\sec\theta\ \sec^2\theta\ d\theta⟶I=4∫secθ sec

2

θ dθ

Performing integration by parts,

\displaystyle\longrightarrow I=4\left[\sec\theta\tan\theta-\int \sec\theta\tan^2\theta\ d\theta\right]⟶I=4[secθtanθ−∫secθtan

2

θ dθ]

\displaystyle\longrightarrow I=4\sec\theta\tan\theta-4\int \sec\theta(\sec^2\theta-1)\ d\theta⟶I=4secθtanθ−4∫secθ(sec

2

θ−1) dθ

\displaystyle\longrightarrow I=4\sec\theta\tan\theta-4\int(\sec^3\theta-\sec\theta)\ d\theta⟶I=4secθtanθ−4∫(sec

3

θ−secθ) dθ

\displaystyle\longrightarrow I=4\sec\theta\tan\theta-4\int\sec^3\theta\ d\theta+4\int\sec\theta\ d\theta\quad\quad\dots(2)⟶I=4secθtanθ−4∫sec

3

θ dθ+4∫secθ dθ…(2)

Adding (1) and (2) we get,

\displaystyle\longrightarrow 2I=4\sec\theta\tan\theta+4\int\sec\theta\ d\theta⟶2I=4secθtanθ+4∫secθ dθ

\displaystyle\longrightarrow I=2\sec\theta\tan\theta+2\log|\sec\theta+\tan\theta|+C'⟶I=2secθtanθ+2log∣secθ+tanθ∣+C

′

Undoing substitutions for \sec\thetasecθ and \tan\theta,tanθ,

\displaystyle\longrightarrow I=\dfrac{1}{2}(x+1)\sqrt{x^2+2x+5}+2\log\left|\dfrac{x+1+\sqrt{x^2+2x+5}}{2}\right|+C'⟶I=

2

1

(x+1)

x

2

+2x+5

+2log

∣

∣

∣

∣

∣

∣

2

x+1+

x

2

+2x+5

∣

∣

∣

∣

∣

∣

+C

′

Taking C'-2\log2=C,C

′

−2log2=C,

\displaystyle\longrightarrow\underline{\underline{I=\dfrac{1}{2}(x+1)\sqrt{x^2+2x+5}+2\log\left|x+1+\sqrt{x^2+2x+5}\right|+C}}⟶

I=

2

1

(x+1)

x

2

+2x+5

+2log

∣

∣

∣

∣

x+1+

x

2

+2x+5

∣

∣

∣

∣

+C

Step-by-step explanation:

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