Solved: "Find the integrals. ∫x2−25xdx"

Dillard

Answered question

2021-10-12

Find the integrals.

\int \frac{\sqrt{x^{2} - 25}}{x} d x

coffentw

Skilled2021-10-13Added 103 answers

Given:

\int \frac{\sqrt{x^{2} - 25}}{x} d x

Explanation:
Let

I = \int \frac{\sqrt{x^{2} - 25}}{x} d x

Put

x = 5 \sec u

\Rightarrow d x = 5 \sec u \tan u d u

I = \int \frac{\sqrt{25 \sec^{2} u - 25}}{5 \sec u} 5 \sec u \tan u d u

= \int \sqrt{25 (\sec^{2} u - 1)} \tan u d u

\int 5 \sqrt{\sec^{2} u - 1} \tan u d u

= 5 \int \sqrt{\sec^{2} u - 1} \tan u d u

= 5 \int \sqrt{\tan^{2} u} \tan u d u

= 5 \int \tan u \tan u d u

= 5 \int \tan^{2} u d u

= 5 \int (\sec^{2} u - 1) d u

= 5 \int \sec^{2} u d u - 5 \int d u

= 5 \tan u - 5 u

= 5 (\sqrt{\sec^{2} u - 1}) - 5 u

= \sqrt{25 \sec^{2} u - 25} - 5 u

= \sqrt{{(5 \sec u)}^{2} - 25} - 5 u

= \sqrt{x^{2} - 25} - 5 \sec^{- 1} (\frac{x}{5}) + C

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