Mathematics

# If $I=\int { \dfrac { 1 }{ { x }^{ 4 }\sqrt { { a }^{ 2 }+{ x }^{ 2 } } } dx, }$ then $I$ equals

$\dfrac { 1 }{ { a }^{ 4 } } =\left\{ \dfrac { 1 }{ x } \sqrt { { a }^{ 2 }+{ x }^{ 2 } } -\dfrac { 1 }{ { 3x }^{ 3 } } \sqrt { { a }^{ 2 }{ +x }^{ 2 } } \right\} +C$

Its FREE, you're just one step away

Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109

#### Realted Questions

Q1 Passage Hard

In calculating a number of integrals we had to use the method of integration by parts several times in succession. The result could be obtained more rapidly and in a more concise form by using the so-called generalized formula for integration by parts.

$\int u(x)\, v(x)dx\, =\, u(x)\, v_{1}(x)\, -\, u^{}(x)v_{2}(x)\, +\, u^{}(x)\, v_{3}(x)\, -\, .\, +\, (-1)^{n\, -\, 1}u^{n\, -\, 1}(x)v_{n}(x)\, -\, (-1)^{n\, -\, 1}$ $\int\, u^{n}(x)v_{n}(x)\, dx$ where $v_{1}(x)\, =\, \int v(x)dx,\, v_{2}(x)\, =\, \int v_{1}(x)\, dx\, ..\, v_{n}(x)\, =\, \int v_{n\, -\, 1}(x) dx$

Of course, we assume that all derivatives and integrals appearing in this formula exist. The use of the generalized formula for integration  by parts is especially useful when calculating $\int P_{n}(x)\, Q(x)\, dx$, where $P_{n}(x)$, is polynomial of degree n and the factor Q(x) is such that it can be integrated successively n + 1 times.

1 Verified Answer | Published on 17th 09, 2020

Q2 Multiple Correct Hard
Let $f(x) = 7\tan^8x+7\tan^6x-3\tan^4x-3\tan^2x$ for all $x\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$, then the correct expression(s) is (are)
• A. $\displaystyle \int_0^{\dfrac{\pi}{4}}xf(x)dx = \dfrac{1}{6}$
• B. $\displaystyle \int_0^{\dfrac{\pi}{4}}xf(x)dx =1$
• C. $\displaystyle \int_0^{\dfrac{\pi}{4}}xf(x)dx = \dfrac{1}{12}$
• D. $\displaystyle \int_0^{\dfrac{\pi}{4}}f(x)dx = 0$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
If $b > a$ and $\displaystyle I = \int_{a}^{b}\frac{dx}{\sqrt{ (x-a)(b-x)}}$ then $I$ equals
• A. $\pi/2$
• B. $3\pi/2$
• C. $2\pi$
• D. $\pi$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
If $\displaystyle A=\int_{0}^{\pi} \dfrac{cos x}{(x+2)^2} \: dx$, then $\displaystyle \int_{0}^{\dfrac{\pi}{2}} \dfrac{\sin 2x}{(x+1)} \: dx$ is equal to
• A. $\dfrac{1}{\pi+2}-A$
• B. ${1}+\dfrac{1}{\pi+2}-A$
• C. $A-\dfrac{1}{2}-\dfrac{1}{\pi+2}$
• D. $\dfrac{1}{2}+\dfrac{1}{\pi+2}-A$

1 Verified Answer | Published on 17th 09, 2020

Q5 Single Correct Medium
The value of $\displaystyle \int_{\sqrt{log2}}^{\sqrt{log3}}\dfrac{xsinx^2}{sinx^2+sin(log6-x^2)}dx$ is
• A. $\dfrac{1}{2}log\dfrac{3}{2}$
• B. $log\dfrac{3}{2}$
• C. $\dfrac{1}{6}log\dfrac{3}{2}$
• D. $\dfrac{1}{4}log\dfrac{3}{2}$