Mathematics

# Solve : $\displaystyle \int \dfrac{u}{v} dx$

##### SOLUTION
Formula for integration by parts:
$\displaystyle\int uvdx=u\displaystyle v dx-\displaystyle\int u'(\displaystyle vdx)dx$
$\displaystyle\int \dfrac{u}{v}dx=\displaystyle\int u\times \left(\dfrac{1}{v}\right)dx$
Applying integration by parts formula
$\therefore \displaystyle\int \dfrac{u}{v}dx=u\displaystyle\int \left(\dfrac{1}{v}\right)dx-\displaystyle\int u'\left(\dfrac{1}{v}dx\right)dx$.

Its FREE, you're just one step away

Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Subjective Medium
Integrate the rational function   $\cfrac {3x-1}{(x-1)(x-2)(x-3)}$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
Let $\displaystyle f\left ( x \right )=\int \frac{x^{2}dx}{\left ( 1+x^{2} \right )\left ( 1+\sqrt{1+x^{2}} \right )}$ and $\displaystyle f\left ( 0 \right )=0.$ Then $\displaystyle f\left ( 1 \right )$ is
• A. $\displaystyle \log \left ( 1+\sqrt{2} \right )$
• B. $\displaystyle \log \left ( 1+\sqrt{2} \right )+\frac{\pi }{4}$
• C. none of these
• D. $\displaystyle \log \left ( 1+\sqrt{2} \right )-\frac{\pi }{4}$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Prove that
$\displaystyle \int _ { 0 } ^ { 1 } \cfrac { 2 n + 3 } { 5 n ^ { 2 } + 1 } d x$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
$\displaystyle \int \frac{\cot^{-1}(e^x)}{e^x} dx$ is equal to:
• A. $\displaystyle \frac{1}{2}\ln (e^{2x}+1)-\displaystyle \frac{\cot^{-1}(e^x)}{(e^x)}+x+c$
• B. $\displaystyle \frac{1}{2}\ln (e^{2x}+1)+\displaystyle \frac{\cot^{-1}(e^x)}{(e^x)}+x+c$
• C. $\displaystyle \frac{1}{2}\ln (e^{2x}+1)+\displaystyle \frac{\cot^{-1}(e^x)}{(e^x)}-x+c$
• D. $\displaystyle \frac{1}{2}\ln (e^{2x}+1)-\displaystyle \frac{\cot^{-1}(e^x)}{(e^x)}-x+c$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020

Q5 Passage Hard
Let us consider the integral of the following forms
$f{(x_1,\sqrt{mx^2+nx+p})}^{\tfrac{1}{2}}$
Case I If $m>0$, then put $\sqrt{mx^2+nx+C}=u\pm x\sqrt{m}$
Case II If $p>0$, then put $\sqrt{mx^2+nx+C}=u\pm \sqrt{p}$
Case III If quadratic equation $mx^2+nx+p=0$ has real roots $\alpha$ and $\beta$, then put $\sqrt{mx^2+nx+p}=(x-\alpha)u\:or\:(x-\beta)u$

Asked in: Mathematics - Integrals

1 Verified Answer | Published on 17th 09, 2020