Mathematics

If $\displaystyle \int { \frac { { x }^{ 2 }-x-1 }{ \sqrt [ 3 ]{ 1-3x } } dx } =\frac { -\left( { x }^{ 2 }-x-1 \right) { \left( 1-3x \right) }^{ \frac { 2 }{ 3 } } }{ 2 } -\frac { \left( 2x-1 \right) { \left( 1-3x \right) }^{ \frac { 8 }{ 3 } } }{ k } +C$, where $k=$

40

SOLUTION
$\displaystyle I=\int { \frac { { x }^{ 2 }-x-1 }{ \sqrt [ 3 ]{ 1-3x } } dx } =\int { \left( { x }^{ 2 }-x-1 \right) { \left( 1-3x \right) }^{ \frac { -1 }{ 3 } } } dx$

$\displaystyle=\left( { x }^{ 2 }-x-1 \right) \frac { { \left( 1-3x \right) }^{ \frac { 2 }{ 3 } } }{ \left( \frac { 2 }{ 3 } \right) \left( -3 \right) }-\int { \frac { { \left( 1-3x \right) }^{ \frac { 2 }{ 3 } } }{ \left( \frac { 2 }{ 3 } \right) \left( -3 \right) } } .\left( 2x-1 \right) dx$

$\displaystyle=\frac { \left( { x }^{ 2 }-x-1 \right) { \left( 1-3x \right) }^{ \frac { 2 }{ 3 } } }{ -2 } +\frac { 1 }{ 2 } \int { \left( 2x-1 \right) } { \left( 1-3x \right) }^{ \frac { 2 }{ 3 } }dx$

$\displaystyle=\frac { \left( { x }^{ 2 }-x-1 \right) { \left( 1-3x \right) }^{ \frac { 2 }{ 3 } } }{ -2 } +\frac { 1 }{ 2 } \left( 2x-1 \right) \frac { { \left( 1-3x \right) }^{ \frac { 5 }{ 3 } } }{ \left( \frac { 5 }{ 3 } \right) \left( -3 \right) }-\frac { 1 }{ 2 } \int { \frac { { \left( 1-3x \right) }^{ \frac { 5 }{ 3 } } }{ \left( \frac { 5 }{ 3 } \right) \left( -3 \right) } }$

$\displaystyle=\frac { \left( { x }^{ 2 }-x-1 \right) { \left( 1-3x \right) }^{ \frac { 2 }{ 3 } } }{ -2 } +\frac { \left( 2x-1 \right) { \left( 1-3x \right) }^{ \frac { 5 }{ 3 } } }{ -10 } +\frac { 1 }{ 5 } \frac { { \left( 1-3x \right) }^{ \frac { 8 }{ 3 } } }{ \left( \frac { 8 }{ 3 } \right) \left( -3 \right) } +C$

$\displaystyle=\frac { \left( { x }^{ 2 }-x-1 \right) { \left( 1-3x \right) }^{ \frac { 2 }{ 3 } } }{ -2 } +\frac { \left( 2x-1 \right) { \left( 1-3x \right) }^{ \frac { 5 }{ 3 } } }{ -10 } +\frac { { \left( 1-3x \right) }^{ \frac { 8 }{ 3 } } }{ -40 } +C$
$\therefore k=40$

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One Word Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

Realted Questions

Q1 Single Correct Medium
If $I=\int _{ 0 }^{ 2\pi }{ { sin }^{ 2 }xdx }$, then
• A. $I=4\int _{ 0 }^{ 2\pi }{ { sin }^{ 2 }xdx }$
• B. $I=2\int _{ 0 }^{ 2\pi }{ { cos }^{ 2 }xdx }$
• C. $I=8\int _{ 0 }^{ 4\pi }{ { sin }^{ 2 }xdx }$
• D. $I=2\int _{ 0 }^{ 2\pi }{ { sin }^{ 2 }xdx }$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
If $\displaystyle \frac{\pi }{4}< \alpha < \displaystyle \frac{\pi }{2}$, value of $\displaystyle \int_{-\pi /2}^{\pi /2}\displaystyle \frac{\sin 2x}{\sqrt{1+\sin 2\alpha \sin x}}$ is
• A. $-\displaystyle \frac{4}{3}\tan \alpha \sec \alpha$
• B. $-\displaystyle \frac{4}{3}\tan \alpha cosec\alpha$
• C. $-\displaystyle \frac{4}{3}\cot \alpha \sec \alpha$
• D. $-\displaystyle \frac{4}{3}\cot \alpha cosec\alpha$

1 Verified Answer | Published on 17th 09, 2020

Q3 Multiple Correct Hard
If $A=\displaystyle \int_{0}^{\pi }\frac{\sin x}{\sin x+\cos x}dx,B=\int_{0}^{\pi }\frac{\sin x}{\sin x-\cos x}dx$ then
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• B. $\displaystyle A=-B=\pi$
• C. $A=B$
• D. $\displaystyle A=B=\pi /2$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
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