Q.Find $\displaystyle\int \sqrt{1+3x-x^2}\, dx$.

Complete the square under the root, then use the standard integral $\int dx/\sqrt{a^2-x^2}=\sin^{-1}(x/a)+c$.

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

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Complete the square in the denominator, then use the standard $\int \frac{dt}{t^2+a^2}$ form.

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

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The expression under the root is a perfect square, so the square root simplifies to a linear expression.

Note that $x^2+3x+\dfrac{9}{4}=\left(x+\dfrac{3}{2}\right)^2$.

So $\sqrt{x^2+3x+\dfrac{9}{4}}=\left|x+\dfrac{3}{2}\right|$, which (taking the positive branch) is $x+\dfrac{3}{2}$.

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This is a standard integral of the form $\int\sqrt{x^2+a^2}\,dx$.

The standard formula (derivable by integration by parts, treating $\sqrt{1+x^2}=\sqrt{1+x^2}\cdot 1$) is:

$\int \sqrt{x^2+a^2}\,dx = \frac{x}{2}\sqrt{x^2+a^2} + \frac{a^2}{2}\log\left|x+\sqrt{x^2+a^2}\right| + c$

With $a=1$: …

Standard result: $\int\sqrt{a^2-x^2}\,dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+c$.

This is a memorised standard form (derivable by the substitution $x=a\sin\theta$):

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With $u=3x$, $du=3\,dx$, the integral becomes $\int\dfrac{du}{\sqrt{1-u^{2}}}=\sin^{-1}3x+k$.

Let $u=3x$, so $du=3\,dx$. The numerator $3\,dx=du$.

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Completing the square, $x^2+2x+2=(x+1)^2+1$, giving $\tan^{-1}(x+1)+C$.

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

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