Advertisements
Advertisements
Question
Integrate the function `(3x)/(1+ 2x^4)`
Solution
Let `I = int (3x)/ (1 + 2x^4) dx`
Put x2 = t
⇒ 2x dx = dt
⇒ `x dx = dt/2`
∴ `I = 3/2 int dt/(1 + 2t^2 )`
`= 3/4 int dt/ (1/2 + t^2)`
`= 3/4 int dt/ ((1/sqrt2)^2 + t^2)` `....[∵ int dx/(a^2+x^2) = 1/a tan^-1 x/a + C]`
`3/4* 1/ (1/sqrt2) tan^-1 (t/(1/sqrt2)) + C`
`3/(2sqrt2) tan^-1 sqrt2t + C`
APPEARS IN
RELATED QUESTIONS
Evaluate: `int(5x-2)/(1+2x+3x^2)dx`
Evaluate : ` int x^2/((x^2+4)(x^2+9))dx`
find : `int(3x+1)sqrt(4-3x-2x^2)dx`
Find:
`int(x^3-1)/(x^3+x)dx`
Evaluate:
`int((x+3)e^x)/((x+5)^3)dx`
Integrate the function `1/sqrt((2-x)^2 + 1)`
Integrate the function `x^2/(1 - x^6)`
Integrate the function `(x - 1)/sqrt(x^2 - 1)`
Integrate the function `1/sqrt((x -1)(x - 2))`
Integrate the function `1/sqrt((x - a)(x - b))`
Integrate the function `(4x+ 1)/sqrt(2x^2 + x - 3)`
Integrate the function `(6x + 7)/sqrt((x - 5)(x - 4))`
Integrate the function `(x + 3)/(x^2 - 2x - 5)`
Integrate the function `(5x + 3)/sqrt(x^2 + 4x + 10)`
`int dx/(x^2 + 2x + 2)` equals:
Integrate the function:
`sqrt(4 - x^2)`
Integrate the function:
`sqrt(x^2 + 4x +1)`
Integrate the function:
`sqrt(1-4x - x^2)`
`int sqrt(x^2 - 8x + 7) dx` is equal to ______.
\[\int\frac{8x + 13}{\sqrt{4x + 7}} \text{ dx }\]
\[\int\frac{1 + x + x^2}{x^2 \left( 1 + x \right)} \text{ dx}\]
Find : \[\int\left( 2x + 5 \right)\sqrt{10 - 4x - 3 x^2}dx\] .
Find:
`int_(-pi/4)^0 (1+tan"x")/(1-tan"x") "dx"`