Advertisements
Advertisements
प्रश्न
`int (x + 3)/(x + 4)^2 "e"^x "d"x` = ______.
Advertisements
उत्तर
`int (x + 3)/(x + 4)^2 "e"^x "d"x` = `"e"^x/(x + 4) + "C"`.
Explanation:
Let I = `int (x + 3)/(x + 4)^2 * "e"^x "d"x`
= `int (x + 4 - 1)/(x + 4)^2 * "e"^x "d"x`
= `int [(x + 4)/(x + 4)^2 - 1/(x + 4)^2]"e"^x "d"x`
= `int [1/(x + 4) - 1/(x + 4)^2]"e"^x "d"x`
Put `1/(x + 4)` = t
⇒ `- 1/(x + 4)^2 "d"x` = dt
Let f(x) = `1/(x + 4)`
∴ f'(x) = `- 1/(x + 4)^2`
Using `int "e"^x ["f"(x) + "f'"(x)]"d"x = "e"^x "f"(x) + "C"`
∴ I = `"e"^x * 1/(x + 4) + "C"`.
APPEARS IN
संबंधित प्रश्न
If \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that \[\int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx\]
If f is an integrable function, show that
If f(x) is a continuous function defined on [−a, a], then prove that
If \[\left[ \cdot \right] and \left\{ \cdot \right\}\] denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:
Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_0^1 \tan^{- 1} x dx\]
\[\int\limits_0^1 \frac{1 - x}{1 + x} dx\]
\[\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx\]
\[\int\limits_0^{\pi/4} e^x \sin x dx\]
\[\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx\]
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
Find : `∫_a^b logx/x` dx
Evaluate the following:
f(x) = `{{:("c"x",", 0 < x < 1),(0",", "otherwise"):}` Find 'c" if `int_0^1 "f"(x) "d"x` = 2
Evaluate the following using properties of definite integral:
`int_(-1)^1 log ((2 - x)/(2 + x)) "d"x`
Evaluate the following integrals as the limit of the sum:
`int_1^3 (2x + 3) "d"x`
Choose the correct alternative:
`int_0^oo "e"^(-2x) "d"x` is
Choose the correct alternative:
`int_0^oo x^4"e"^-x "d"x` is
Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`
`int x^3/(x + 1)` is equal to ______.
