Get Answers to all your Questions

header-bg qa

Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise18.5 Question 4 Maths Textbook Solution.

Answers (1)

Answer:  \frac{2}{135}(9 x+20)(3 x+5)^{\frac{3}{2}}+C

Hint: Let t= 3x+5

Given: \int \left ( x+2 \right )\sqrt{3x+5}dx


Substitute 3x+5=t

\begin{aligned} &\Rightarrow x=\frac{t-5}{3} \Rightarrow 3 d x=d t \\ &\Rightarrow d x=\frac{d t}{3} \end{aligned}

\begin{aligned} &\therefore I=\int\left(\frac{t-5}{3}+2\right) \sqrt{t} \frac{d t}{3} \\ &\therefore I=\frac{1}{3} \int\left(\frac{t-5+6}{3}\right) \sqrt{t} d t \end{aligned}


On taking 3 as common and multiply, we get


=\frac{1}{9} \int\left(t^{\frac{3}{2}}+t^{\frac{1}{2}}\right) d t

On integrating we get


On simplifying

=\frac{1}{9}\left[\frac{2}{5} t^{\frac{5}{2}}+\frac{2}{3} t^{\frac{3}{2}}\right]+C

On substituting the value of t
\begin{aligned} &=\frac{1}{9}\left[\frac{2}{5}(3 x+5)^{\frac{5}{2}}+\frac{2}{3}(3 x+5)^{\frac{3}{2}}\right]+C \\ &=\frac{2}{9}\left[(3 x+5)^{\frac{3}{2}}\left\{\frac{3 x+5}{5}+\frac{1}{3}\right\}\right]+C \end{aligned}

\begin{aligned} &=\frac{2}{9}\left[(3 x+5)^{\frac{3}{2}}\left\{\frac{9 x+15+5}{15}\right\}\right]+C \\ &=\frac{2}{9}\left[(3 x+5)^{\frac{3}{2}}\left\{\frac{9 x+20}{15}\right\}\right]+C \\ &=\frac{2}{135}(3 x+5)^{\frac{3}{2}}(9 x+20)+C \end{aligned}

Posted by


View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support