Mathematics

# Solve :$\int (3x+5)^7 dx$

##### SOLUTION
$\begin{array}{l} \int _{ }^{ }{ { { \left( { 3x+5 } \right) }^{ 7 } } } dx \\ Put\, \, \left( { 3x+5 } \right) =t\, so\, that\, \, 3dx\, =dt\, or\, dx=\frac { 1 }{ 3 } dt. \\ \therefore \int _{ }^{ }{ { { \left( { 3x+5 } \right) }^{ 7 } } } dx \\ =\frac { 1 }{ 3 } .\frac { { { t^{ 8 } } } }{ 8 } +C \\ =\frac { { { { \left( { 3x+5 } \right) }^{ 8 } } } }{ { 24 } } +C \end{array}$

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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 Hard
$\displaystyle \overset{\pi/4}{\underset{0}{\int}}\dfrac{\sin x + \cos x}{9 + 16 \sin 2x} dx$ is equal to
• A. $log \,3$
• B. $log \,2$
• C. $\dfrac{1}{20}log \,3$
• D. $\dfrac{1}{40}log \,9$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\int {x{{\sin }^{ - 1}}xdx}$=?
• A. $\frac{1}{4}{\sin ^{ - 1}}(x)(2x + 1) + \frac{{x\sqrt {1 - {x^2}} }}{4} + c$
• B. $\frac{1}{4}{\cos ^{ - 1}}(x)(2x - 1) + \frac{{x\sqrt {1 - {x^2}} }}{4} + c$
• C. $\frac{1}{4}{\sin ^{ - 1}}(x)(2x - 1) - \frac{{x\sqrt {1 - {x^2}} }}{4} + c$
• D. $\frac{1}{4}{\sin ^{ - 1}}(x)(2x^2 - 1) + \frac{{x\sqrt {1 - {x^2}} }}{4} + c$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
Let $f(x)=3x^{2}-7x+a, x > \dfrac{7}{6}$, the value of a such that $f(x)$ touches its inverse $f^{-1}(x)$ is
• A. $3$
• B. $-3$
• C. $\dfrac{49}{12}$
• D. $\dfrac{16}{3}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
For a function $f,f\left ( a+b-x \right )= f\left ( x \right )$, and $\displaystyle \int_{a}^{b}xf\left ( x \right )dx= k\int_{a}^{b}f\left ( x \right )dx$, then value of $k$ is
• A. $\displaystyle \frac{1}{2}\left ( a-b \right )$
• B. $\displaystyle \frac{1}{2}\left ( a^{2}+b^{2} \right )$
• C. $\displaystyle \frac{1}{2}\left ( a^{2}-b^{2} \right )$
• D. $\displaystyle \frac{1}{2}\left ( a+b \right )$

1 Verified Answer | Published on 17th 09, 2020

Q5 Passage Hard

In calculating a number of integrals we had to use the method of integration by parts several times in succession. The result could be obtained more rapidly and in a more concise form by using the so-called generalized formula for integration by parts.

$\int u(x)\, v(x)dx\, =\, u(x)\, v_{1}(x)\, -\, u^{}(x)v_{2}(x)\, +\, u^{}(x)\, v_{3}(x)\, -\, .\, +\, (-1)^{n\, -\, 1}u^{n\, -\, 1}(x)v_{n}(x)\, -\, (-1)^{n\, -\, 1}$ $\int\, u^{n}(x)v_{n}(x)\, dx$ where $v_{1}(x)\, =\, \int v(x)dx,\, v_{2}(x)\, =\, \int v_{1}(x)\, dx\, ..\, v_{n}(x)\, =\, \int v_{n\, -\, 1}(x) dx$

Of course, we assume that all derivatives and integrals appearing in this formula exist. The use of the generalized formula for integration  by parts is especially useful when calculating $\int P_{n}(x)\, Q(x)\, dx$, where $P_{n}(x)$, is polynomial of degree n and the factor Q(x) is such that it can be integrated successively n + 1 times.