Q.Evaluate the integral using substitution $\int_{-1}^{1}\frac{dx}{x^2+2x+5}$

Complete the square under the root and use $\int\frac{du}{\sqrt{a^2-u^2}}=\sin^{-1}\frac{u}{a}$; result $\frac12\sin^{-1}\frac{8x-9}{9}$.

Complete the square inside the square root and reduce to the standard form $\int dx/\sqrt{x^2-a^2}$.

$5x^2-2x=5\left[\left(x-\dfrac15\right)^2-\dfrac{1}{25}\right]$, so

$\displaystyle\int\frac{dx}{\sqrt{5x^2-2x}}=\frac{1}{\sqrt5}\int\frac{dx}{\sqrt{(x-1/5)^2-(1/5)^2}}=\frac{1}{\sqrt5}\ln\left|\left(x-\frac15\right)+\sqrt{(x-1/5)^2-1/25}\right|+C$

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Complete the square under the root: $4x-9x^2=9\big[(\tfrac29)^2-(x-\tfrac29)^2\big]$. …

Complete the square: $4x-x^2=4-(x-2)^2$, then use $\int\frac{dx}{\sqrt{a^2-u^2}}=\sin^{-1}\frac{u}{a}$.

Complete the square in the denominator to get the standard $\frac{1}{x^2+a^2}$ form.

$x^2+2x+5=(x+1)^2+4$.

$\int\frac{dx}{(x+1)^2+2^2}=\frac{1}{2}\tan^{-1}\left(\frac{x+1}{2}\right)+C$.

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Complete the square under the root and use $\int \tfrac{dx}{\sqrt{t^2-a^2}} = \log|t+\sqrt{t^2-a^2}|$.

$(x-1)(x-2) = x^2-3x+2 = \left(x-\tfrac32\right)^2 - \tfrac14.$

With $t = x-\tfrac32$: …

Complete the square and use $\int\sqrt{t^2-a^2}\,dt = \tfrac{t}{2}\sqrt{t^2-a^2} - \tfrac{a^2}{2}\log|t+\sqrt{t^2-a^2}|$.

$x^2+4x+1 = (x+2)^2 - 3$, so $t = x+2$, $a^2 = 3$.

$\int\sqrt{t^2-3}\,dt = \dfrac{t}{2}\sqrt{t^2-3} - \dfrac{3}{2}\log|t+\sqrt{t^2-3}|.$

Substituting back: …

Complete the square under the root to get the standard form $\int\dfrac{dx}{\sqrt{a^2-u^2}}=\sin^{-1}\dfrac{u}{a}+C$.

$2x-x^2=-(x^2-2x)=-(x^2-2x+1-1)=1-(x-1)^2$.

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Apply the standard formula $\int\sqrt{a^2-x^2}\,dx = \frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C$.

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