Advertisements
Advertisements
Question
`int x^9/(4x^2 + 1)^6 "d"x` is equal to ______.
Options
`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"`
Advertisements
Solution
`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
RELATED QUESTIONS
Evaluate the following integral:
If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]
Evaluate :
\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\] equals
The value of the integral \[\int\limits_{- 2}^2 \left| 1 - x^2 \right| dx\] is ________ .
Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .
Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
\[\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\]
\[\int\limits_0^{\pi/2} \frac{\cos^2 x}{\sin x + \cos x} dx\]
\[\int\limits_0^\pi \cos 2x \log \sin x dx\]
\[\int\limits_0^2 \left( 2 x^2 + 3 \right) dx\]
\[\int\limits_2^3 e^{- x} dx\]
Choose the correct alternative:
Γ(1) is
Find `int x^2/(x^4 + 3x^2 + 2) "d"x`
Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`
Evaluate `int (x^2 + x)/(x^4 - 9) "d"x`
