Mathematics

$\int { \cfrac { f'(x) }{ f(x) } dx } =\log { [f(x)] } +c$

SOLUTION
Let $f(x) =t$

Differentiate on both sides ;

$\Rightarrow f^{'} (x) dx=dt$

Integral becomes : $\dfrac {dt} {t} =log[t] +c$

$=log[f(x)] +c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111

Realted Questions

Q1 Single Correct Medium
$\displaystyle \int^{2}_{1} \dfrac{\log x}{x^{2}}.dx$
• A. $\dfrac{\log 2}{2}+\dfrac{1}{2}$
• B. $\dfrac{-\log 2}{2}-\dfrac{1}{2}$
• C. None of these
• D. $\dfrac{-\log 2}{2}+\dfrac{1}{2}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
Evaluate the following integrals:$\displaystyle \int \sqrt{3x^{2}+4}dx$
• A. $\dfrac{x}{2}.\sqrt{3x+4}+\dfrac{2}{\sqrt{3}} \log\left | \sqrt{3}x+\sqrt{3x^{2}+4} \right |+C$
• B. $\dfrac{x}{2}.\sqrt{3x^{2}+4}+\dfrac{4}{\sqrt{3}} \log\left | \sqrt{3}x+\sqrt{3x^{2}+4} \right |+C$
• C. None of these
• D. $\dfrac{x}{2}.\sqrt{3x^{2}+4}+\dfrac{2}{\sqrt{3}} \log\left | \sqrt{3}x+\sqrt{3x^{2}+4} \right |+C$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Hard
Evaluate: $\displaystyle \int_{0}^{\frac{\pi}{2}} \dfrac{\sin x \cdot \cos x}{1+\sin^4 x}\cdot dx$.

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
Solve: $\displaystyle \int_{}^{} {\frac{{12 + 3}}{{t + 1}}} \,\,dt$
• A. $-15\ln|t+1|+c$
• B. $15\ln|t-1|+c$
• C. $-15\ln|t-1|+c$
• D. $15\ln|t+1|+c$

Consider two differentiable functions $f(x), g(x)$ satisfying $\displaystyle 6\int f(x)g(x)dx=x^{6}+3x^{4}+3x^{2}+c$ & $\displaystyle 2 \int \frac {g(x)dx}{f(x)}=x^{2}+c$. where $\displaystyle f(x)>0 \forall x \in R$