Mathematics

# Solve:$\displaystyle \int \dfrac{dx}{\sqrt{x+a}}dx$

##### SOLUTION
$\displaystyle \int \dfrac {1}{\sqrt {x+4}} dx$
Let $x+e=t^1$
$dx=2+dt$
$\displaystyle \int \dfrac {2t}{t}dt \ \Rightarrow \ 2t$
$\Rightarrow \ 2\sqrt {x+a}+c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 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

Q2 Subjective Medium
Integrate:
$\int \dfrac{x+2^x.\log 2}{x^2+2^{x+1}}dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Hard
Evaluate : $\displaystyle\int \dfrac {x^{2} - 1}{(x - 1)^{2}(x + 3)}dx$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
$\displaystyle\int\frac{dx}{\sin x-\cos x+\sqrt{2}}$ is equal to.
• A. $-\displaystyle\frac{1}{\sqrt{2}}\tan\left(\displaystyle\frac{x}{2}+\frac{\pi}{8}\right)+C$
• B. $\displaystyle\frac{1}{\sqrt{2}}\tan\left(\displaystyle\frac{x}{2}+\frac{\pi}{8}\right)+C$
• C. $\displaystyle\frac{1}{\sqrt{2}}\cot\left(\displaystyle\frac{x}{2}+\frac{\pi}{8}\right)+C$
• D. $-\displaystyle\frac{1}{\sqrt{2}}\cot\left(\displaystyle\frac{x}{2}+\frac{\pi}{8}\right)+C$

Let $\displaystyle f\left ( x \right )=\frac{\sin 2x \cdot \sin \left ( \dfrac{\pi }{2}\cos x \right )}{2x-\pi }$