Advertisements
Advertisements
प्रश्न
Evaluate `int (x^2"d"x)/(x^4 + x^2 - 2)`
Advertisements
उत्तर
Let x2 = t.
Then `x^2/(x^4 + x^2 - 2) = "t"/("t"^2 + "t" - 2)`
= `"t"/(("t" + 2)("t" - 1))`
= `"A"/("t" + 2) + "B"/("t" - 1)`
So t = A(t – 1) + B(t + 2)
Comparing coefficients, we get A = `2/3`, B = `1/3`.
So `x^2/(x^4 + x^2 - 2) = 2/3 1/(x^2 + 2) + 1/3 1/(x^2 - 1)`
Therefore, `int x^2/(x^4 + x^2 - 2) "d"x`
= `2/3 int 1/(x^2 + 2) "d"x + 1/3 int "dx"/(x^2 - 1)`
= `2/3 1/sqrt(2) tan^-1 x/sqrt(2) + 1/6 log |(x + 1)/(x + 1)| + "C"`
APPEARS IN
संबंधित प्रश्न
Evaluate the following definite integrals:
If f is an integrable function, show that
If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]
\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.
Evaluate : \[\int e^{2x} \cdot \sin \left( 3x + 1 \right) dx\] .
\[\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx\]
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
\[\int\limits_1^4 \left( x^2 + x \right) dx\]
\[\int\limits_0^3 \left( x^2 + 1 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_1^"e" ("d"x)/(x(1 + logx)^3`
