Expert answers to "Integrals using trigonometric substitution ∫dxx2+4x"

melodykap

Answered question

2021-10-20

Integrals using trigonometric substitution

\int \frac{d x}{\sqrt{x^{2} + 4 x}}

svartmaleJ

Skilled2021-10-21Added 92 answers

Given:

I = \int \frac{\sin (\ln x)}{4 x} d x

for evaluating given integral, we make denominator in the form of

\sqrt{{(x + a)}^{2} - b^{2}}

so,

\int \frac{d x}{\sqrt{x^{2} + 4 x}} = \int \frac{d x}{\sqrt{x^{2} + 2 (x) (2) + 2^{2} - 2^{2}}}

= \int \frac{d x}{\sqrt{{(x + 2)}^{2} - 2^{2}}}

now we know that

\int \frac{d x}{\sqrt{{(x + a)}^{2} - b^{2}}} = \log | (x + a) + \sqrt{{(x + a)}^{2} - b^{2}} | + c

so,

\int \frac{d x}{\sqrt{{(x + 2)}^{2} - 2^{2}}} = \log | (x + 2) + \sqrt{{(x + 2)}^{2} - 2^{2}} | + c

= \log | (x + 2) + \sqrt{x^{2} + 4 x} | + c

hence, given integral is equal to

\log | (x + 2) + \sqrt{x^{2} + 4 x} | + c

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