Advertisements
Advertisements
प्रश्न
Evaluate the following:
`int (3x - 1)/sqrt(x^2 + 9) "d"x`
Advertisements
उत्तर
Let I = `int (3x - 1)/sqrt(x^2 + 9) "d"x`
= `int (3x)/sqrt(x^2 + 9) "d"x - int 1/sqrt(x^2 + 9) "d"x`
I = I1 – I2
Now I1 = `int (3x)/sqrt(x^2 + 9) "d"x`
Put x2 + 9 = t
⇒ 2x dx = dt
x dx = – dt
∴ I1 = `3/2 int "dt"/sqrt("t")`
= `3/2 * 2sqrt("t") + "C"_1`
= `3sqrt(x^2 + 9) + "C"_1`
I2 = `int 1/sqrt(x^2 + 9) "d"x`
= `int 1/sqrt(x^2 + (3)^2) "d"x`
= `log|x + sqrt(x^2 + (3)^2)| + "C"_2` ....`[because int 1/sqrt(x^2 + "a"^2) "d"x = log|x + sqrt(x^2 + "a"^2)| + "C"]`
= `log|x + sqrt(x^2 + 9)| + "C"_2`
∴ I = I1 – I2
= `3sqrt(x^2 + 9) + "C"_1 - log|x + sqrt(x^2 + 9)| - "C"_2`
= `3sqrt(x^2 + 9) - log|x + sqrt(x^2 + 9)| + ("C"_1 - "C"_2)`
Hence, I = `3sqrt(x^2 + 9) - log|x + sqrt(x^2 + 9)| + "C"`
APPEARS IN
संबंधित प्रश्न
\[\int\frac{\left\{ e^{\sin^{- 1} }x \right\}^2}{\sqrt{1 - x^2}} dx\]
\[\int\frac{\cot x}{\sqrt{\sin x}} dx\]
Evaluate the following integrals:
Evaluate the following integral :-
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate:\[\int\frac{x^2}{1 + x^3} \text{ dx }\] .
Evaluate:\[\int\frac{\sec^2 \sqrt{x}}{\sqrt{x}} \text{ dx }\]
Evaluate:\[\int\frac{\left( 1 + \log x \right)^2}{x} \text{ dx }\]
Evaluate:\[\int \sec^2 \left( 7 - 4x \right) \text{ dx }\]
Evaluate:\[\int\frac{\log x}{x} \text{ dx }\]
Evaluate: \[\int 2^x \text{ dx }\]
Evaluate:\[\int\frac{e\tan^{- 1} x}{1 + x^2} \text{ dx }\]
Evaluate: \[\int\frac{x + \cos6x}{3 x^2 + \sin6x}\text{ dx }\]
Evaluate : \[\int\frac{1}{x(1 + \log x)} \text{ dx}\]
Evaluate: `int_ (x + sin x)/(1 + cos x ) dx`
Evaluate the following:
`int sqrt(5 - 2x + x^2) "d"x`
Evaluate the following:
`int sqrt(2"a"x - x^2) "d"x`
