A man plans to take a housing loan of Rs 99,53,000 from a bank costing 18% per annum compounded monthly. The loan is to be paid back in 30 years in…

The integral $\displaystyle \int \frac{dx}{\sqrt{5 - 4x - x^2}}$∫5−4x−x2​dx​ requires converting the expression under the square root into standard form.

The standard formula needed is:

$$\int \frac{dx}{\sqrt{a^2 - u^2}} = \sin^{-1}\left(\frac{u}{a}\right) + C$$∫a2−u2​dx​=sin−1(au​)+C

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To convert $5 - 4x - x^2$5−4x−x2 into the form $a^2 - u^2$a2−u2, complete the square.

Taking out the negative sign from the $x^2$x2 and $x$x terms:

$$5 - 4x - x^2 = -(x^2 + 4x) + 5$$5−4x−x2=−(x2+4x)+5

Adding and subtracting $\left(\frac{4}{2}\right)^2 = 4$(24​)2=4 inside the bracket:

$$= -(x^2 + 4x + 4 - 4) + 5$$=−(x2+4x+4−4)+5

$$= -[(x+2)^2 - 4] + 5$$=−[(x+2)2−4]+5

$$= -(x+2)^2 + 4 + 5$$=−(x+2)2+4+5

$$= 9 - (x+2)^2$$=9−(x+2)2

$$= 3^2 - (x+2)^2$$=32−(x+2)2

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The integral becomes:

$$\int \frac{dx}{\sqrt{9 - (x+2)^2}}$$∫9−(x+2)2​dx​

This matches the standard form with $a = 3$a=3 and $u = (x+2)$u=(x+2).

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Applying the standard formula:

$$\int \frac{dx}{\sqrt{9 - (x+2)^2}} = \sin^{-1}\left(\frac{x+2}{3}\right) + C$$∫9−(x+2)2​dx​=sin−1(3x+2​)+C

Therefore, the answer is $\sin^{-1}\left(\frac{x + 2}{3}\right) + C$sin−1(3x+2​)+C