English

Ed∫x+3(x+4)2ex dx = ______.

Advertisements
Advertisements

Question

`int (x + 3)/(x + 4)^2 "e"^x  "d"x` = ______.

Fill in the Blanks
Advertisements

Solution

`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"`.

shaalaa.com
  Is there an error in this question or solution?
Chapter 7: Integrals - Exercise [Page 169]

APPEARS IN

NCERT Exemplar Mathematics Exemplar [English] Class 12
Chapter 7 Integrals
Exercise | Q 60 | Page 169

RELATED QUESTIONS

\[\int\limits_2^3 \frac{x}{x^2 + 1} dx\]

\[\int\limits_{- 1}^1 \frac{1}{1 + x^2} dx\]

\[\int\limits_0^{\pi/2} \cos^4\ x\ dx\]

 


\[\int\limits_0^1 \frac{1}{2 x^2 + x + 1} dx\]

\[\int\limits_0^{2\pi} e^{x/2} \sin\left( \frac{x}{2} + \frac{\pi}{4} \right) dx\]

\[\int_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\]

\[\int\limits_0^1 \frac{2x}{1 + x^4} dx\]

\[\int\limits_0^{\pi/2} \frac{1}{a^2 \sin^2 x + b^2 \cos^2 x} dx\]

\[\int_{- 1}^2 \left( \left| x + 1 \right| + \left| x \right| + \left| x - 1 \right| \right)dx\]

 


If  \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that \[\int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx\]

 


\[\int\limits_0^{\pi/2} \frac{\sin^{3/2} x}{\sin^{3/2} x + \cos^{3/2} x} dx\]

\[\int\limits_0^4 \left( x + e^{2x} \right) dx\]

\[\int\limits_2^3 x^2 dx\]

\[\int\limits_{- \pi/2}^{\pi/2} \cos^2 x\ dx .\]

\[\int\limits_2^3 \frac{1}{x}dx\]

\[\int\limits_0^{15} \left[ x \right] dx .\]

\[\int\limits_0^1 \sqrt{x \left( 1 - x \right)} dx\] equals

\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\]  equals


\[\int\limits_0^\pi \frac{1}{a + b \cos x} dx =\]

The value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is 

 


\[\lim_{n \to \infty} \left\{ \frac{1}{2n + 1} + \frac{1}{2n + 2} + . . . + \frac{1}{2n + n} \right\}\] is equal to

Evaluate : \[\int e^{2x} \cdot \sin \left( 3x + 1 \right) dx\] .


\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]


\[\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx\]


\[\int\limits_0^{\pi/2} \frac{\cos^2 x}{\sin x + \cos x} dx\]


\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]


\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]


\[\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx\]


\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]


Prove that `int_a^b ƒ ("x") d"x" = int_a^bƒ(a + b - "x") d"x" and "hence evaluate" int_(π/6)^(π/3) (d"x")/(1+sqrt(tan "x")`


Evaluate the following using properties of definite integral:

`int_0^(i/2) (sin^7x)/(sin^7x + cos^7x)  "d"x`


Evaluate the following:

`int_0^oo "e"^(- x/2) x^5  "d"x`


Choose the correct alternative:

The value of `int_(- pi/2)^(pi/2) cos  x  "d"x` is


Choose the correct alternative:

If n > 0, then Γ(n) is


Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`


Evaluate `int sqrt((1 + x)/(1 - x)) "d"x`, x ≠1


If x = `int_0^y "dt"/sqrt(1 + 9"t"^2)` and `("d"^2y)/("d"x^2)` = ay, then a equal to ______.


`int "e"^x ((1 - x)/(1 + x^2))^2  "d"x` is equal to ______.


Which notation represents a definite integral?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×