Advertisements
Advertisements
Question
Integrate the function `(3x^2)/(x^6 + 1)`
Advertisements
Solution
Let `I = int (3x^2)/(x^6 + 1) dx`
x3 = t Substituting,
3x2 dx = dt
Hence, `I = int 1/(t^2 + 1) dt ... (because d/dx tan^-1 x = 1/(1 + x^2))`
= tan-1 t + C
= tan-1 x3 + C
APPEARS IN
RELATED QUESTIONS
Integrate the function `1/sqrt((2-x)^2 + 1)`
Integrate the function `(3x)/(1+ 2x^4)`
Integrate the function `(x - 1)/sqrt(x^2 - 1)`
Integrate the function `x^2/sqrt(x^6 + a^6)`
Integrate the function `(sec^2 x)/sqrt(tan^2 x + 4)`
Integrate the function `1/sqrt(x^2 +2x + 2)`
Integrate the function `1/sqrt(7 - 6x - x^2)`
Integrate the function `1/sqrt((x -1)(x - 2))`
Integrate the function `1/sqrt(8+3x - x^2)`
Integrate the function `(x + 2)/sqrt(x^2 -1)`
Integrate the function `(x + 2)/sqrt(4x - x^2)`
Integrate the function `(5x + 3)/sqrt(x^2 + 4x + 10)`
Integrate the function:
`sqrt(x^2 + 4x +1)`
Integrate the function:
`sqrt(1-4x - x^2)`
Integrate the function:
`sqrt(x^2 + 4x - 5)`
Integrate the function:
`sqrt(1+ 3x - x^2)`
Integrate the function:
`sqrt(x^2 + 3x)`
Integrate the function:
`sqrt(1+ x^2/9)`
Find `int dx/(5 - 8x - x^2)`
\[\int\frac{8x + 13}{\sqrt{4x + 7}} \text{ dx }\]
\[\int\frac{1 + x + x^2}{x^2 \left( 1 + x \right)} \text{ dx}\]
Find `int (dx)/sqrt(4x - x^2)`
Find: `int (dx)/(x^2 - 6x + 13)`
`int (a^x - b^x)^2/(a^xb^x)dx` equals ______.
