Advertisements
Advertisements
प्रश्न
`int x^9/(4x^2 + 1)^6 "d"x` is equal to ______.
विकल्प
`1/(5x)(4 + 1/x^2)^-5 + "C"`
`1/5(4 + 1/x^2)^-5 + "C"`
`1/(10x)(1 + 4)^-5 + "C"`
`1/10(1/x^2 + 4)^-5 + "C"`
उत्तर
`int x^9/(4x^2 + 1)^6 "d"x` is equal to `1/10(1/x^2 + 4)^-5 + "C"`.
Explanation:
Let I = `int x^9/(4x^2 + 1)^6 "d"x`
= `int x^9/(x^12(4 + 1/x^2)^6) "d"x`
= `int 1/(x^3(4 + 1/x^2)^6) "d"x`
Put `(4 + 1/x^2)` = t
⇒ `(-2)/x^3 "dt"` = dt
⇒ `"dx"/x^3 = - 1/2 "dt"`
∴ I = `- 1/2 int "dt"/"t"^6`
= `- 1/2 xx - 1/5 "t"^-5 + "C"`
= `1/10 "t"^-5 + "C"`
= `1/10(4 + 1/x^2)^-5 + "C"`
APPEARS IN
संबंधित प्रश्न
Evaluate each of the following integral:
\[\int_a^b \frac{x^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx, n \in N, n \geq 2\]
Evaluate the following integral:
Evaluate each of the following integral:
The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
\[\int\limits_0^{2\pi} \cos^7 x dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]
\[\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_0^\pi \cos 2x \log \sin x dx\]
\[\int\limits_0^{\pi/2} \frac{dx}{4 \cos x + 2 \sin x}dx\]
Evaluate the following integrals as the limit of the sum:
`int_1^3 (2x + 3) "d"x`
Evaluate the following integrals as the limit of the sum:
`int_0^1 x^2 "d"x`
If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.
Evaluate the following:
`int ((x^2 + 2))/(x + 1) "d"x`
If `intx^3/sqrt(1 + x^2) "d"x = "a"(1 + x^2)^(3/2) + "b"sqrt(1 + x^2) + "C"`, then ______.
