Question

Find the following integrals :-

$(1). \int \frac{dx}{ {x}^{2} - 16 } \\ \\ (2).\int \frac{dx}{ \sqrt{2x - {x}^{2} } }$
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Answer 1

Answer:

Given

a property dealer charges 2% commission from a buyer or seller of property

Mr Ramlal paid Rs20000 as a Commission for purchasing a flat.

What was  the value of the flat.

To proof

Let us assume that the value of the flat be x .

2 % written in the dcimal form

= 0.02

than the equation becomes

x × 0.02 = 20000

x = Rs 1000000

therefore the value of the flat is  Rs 1000000 .

Hence proved

Step-by-step explanation:

Answer 2

Question :-

Find the following integrals:

$(1). \int \frac{dx}{ {x}^{2} - 16 } \\ \\ (2).\int \frac{dx}{ \sqrt{2x - {x}^{2} } }$

Solution :-

$(1). \: we \: \: have \: \: \int \frac{dx}{ {x}^{2} - 16 } = \int \frac{dx}{ {x}^{2} - {4}^{2} } = \frac{1}{8} \: log \: | \frac{x - 4}{x + 4} | + c\\ \\(2). \: \int \frac{dx}{ \sqrt{2x - {x}^{2} } } = \int \frac{dx}{ \sqrt{1 - (x - 1) ^{2} } } \\ \\ put \: x - 1 = t. \: \: Then \: \: dx \: = dt. \\ \\ Therefore, \: \: \: \int \frac{dx}{ \sqrt{2x - {x}^{2} } } = \int \frac{dt}{1 - {t}^{2} } = sin ^{ - 1} \: (t) + c \\ \\ = \: sin^{ - 1} \: (x - 1) + c$