Mathematics

# $\displaystyle \int\left( {{x^3} - 1} \right)dx$

##### SOLUTION
$\displaystyle \int a^3-1=\dfrac {x^4}{4}-x+c$

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

#### Realted Questions

Q1 Single Correct Medium
Evaluate $\displaystyle\int \frac{x^{2}}{9 + 16 x^{6}} dx$
• A. $\displaystyle \frac{1}{16} \tan^{-1} \left (\displaystyle \frac{4x^{3}}{3} \right) + c$
• B. $\displaystyle \frac{1}{36} \tan^{-1} \left (\displaystyle \frac{3x^{3}}{4} \right) + c$
• C. $\displaystyle \frac{1}{16} \tan^{-1} \left (\displaystyle \frac{3x^{3}}{4} \right) + c$
• D. $\displaystyle \frac{1}{36} \tan^{-1} \left (\displaystyle \frac{4x^{3}}{3} \right) + c$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Evaluate: $\displaystyle \int { \dfrac { { x }^{ 3 }-{ x }^{ 2 }+x-1 }{ x-1 } dx }$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Solve: $\int \sin^32x.dx$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard
Solve :
$\displaystyle \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cfrac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x$

1 Verified Answer | Published on 17th 09, 2020

Q5 Single Correct Medium
$\displaystyle\int^{\pi/4}_0(\sqrt{\tan x}+\sqrt{\cot x})dx$ is equal to?
• A. $\dfrac{\pi}{2}$
• B. $-\dfrac{\pi}{2}$
• C. None of the above
• D. $\dfrac{\pi}{\sqrt{2}}$