Mathematics

# Evaluate :     $I=\int { \left( 2x+3 \right) } \sqrt { x-1 } dx$

##### SOLUTION
$I = \int (2x+3)\sqrt{(x-1)}dx$
$= \int (2x-2+5)\sqrt{(x-1)}dx$
$= \int 2(x-1)\sqrt{(x-1)} dx+5 \int \sqrt{(x-1)}dx$
$= 2\int (x-1)^{\frac{3}{2}}dx+5\int (x-1)^{\frac{1}{2}}dx$
$= 2\frac{(x-1)^{\frac{3}{2}+1}}{(\frac{3}{2}+1)}+5\frac{(x-1)^{\frac{1}{2}+1}}{\frac{1}{2}+1} +c$
$I = \frac{4}{5}(x-1)^{^{5/2}}+\frac{10}{3}(x-1)^{\frac{3}{2}}+c$
$\int (2x+3)\sqrt{x-1}dx = (x-1)[\frac{4}{5}(x-1)^{\frac{3}{2}}+\frac{10}{3}\sqrt{x-1}]+c$

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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
The value of $\displaystyle \int \dfrac{x}{\sqrt[3]{a+bx}}dx$ is
• A. $\displaystyle -\dfrac{3}{5b^{2}}\left ( a+bx \right )^{\tfrac{5}{3}}+\frac{3a}{2b^{2}}\left ( a+bx \right )^{\tfrac{2}{3}}+C$
• B. $\displaystyle -\dfrac{3}{5b^{2}}\left ( a+bx \right )^{\tfrac{5}{3}}-\dfrac{3a}{2b^{2}}\left ( a+bx \right )^{\tfrac{2}{3}}+C$
• C. $\displaystyle \dfrac{5}{3b^{2}}\left ( a+bx \right )^{\tfrac{5}{3}}+\dfrac{3a}{2b^{2}}\left ( a+bx \right )^{\tfrac{2}{3}}+C$
• D. $\displaystyle \dfrac{3}{5b^{2}}\left ( a+bx \right )^{\tfrac{5}{3}}-\dfrac{3a}{2b^{2}}\left ( a+bx \right )^{\tfrac{2}{3}}+C$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
$\displaystyle \int \dfrac {2x+5}{x^2+5x+6}dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Hard
Find: $\displaystyle \int{\dfrac{e^{x}(x-3)}{(x-1)^{3}}dx}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Evaluate $\displaystyle \int _2^3 (x^2+2x+5)\ dx$

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 $\displaystyle f(x)>0 \forall x \in R$