#### Please Solve RD Sharma Class 12 Chapter Indefinite Integrals Exercise 18.17 Question 4 Maths Textbook Solution

Answer$\frac{1}{\sqrt{3}}\log \left | \left ( x+\frac{5}{6} \right ) +\sqrt{x^{2}+\frac{5x}{3}}+\frac{7}{3}\right |+c$

Hint:-To solve this problem, use special integration formula.

Given:- $\int \frac{1}{\sqrt{3x^{2}+5x+7}}dx$

Solution:-

Let

$\int \frac{1}{\sqrt{3x^{2}+5x+7}}dx\Rightarrow \int \frac{1}{\sqrt{3\left \{ x^{2}+\frac{5x}{3}+\frac{7}{3} \right \}}}dx$

$\frac{1}{\sqrt{3}}\int \frac{1}{\sqrt{x^{2}+2.x.\frac{5}{6}+\left ( \frac{5}{6} \right )^{2}-\left ( \frac{5}{6} \right )^{2}+\frac{7}{3}}}dx$

$\frac{1}{\sqrt{3}}\int \frac{1}{\sqrt{\left ( x+\frac{5}{6} \right )^{2}-\frac{25}{36}+\frac{7}{3}}}dx$

$\frac{1}{\sqrt{3}}\int \frac{1}{\sqrt{\left ( x+\frac{5}{6} \right )^{2}-\left ( \frac{25-84}{36} \right )}}dx$

$\frac{1}{\sqrt{3}}\int \frac{1}{\sqrt{\left ( x+\frac{5}{6} \right )^{2}+\frac{59}{36}}}dx$

Put

$x+\frac{5}{6}=t\Rightarrow dx=dt$

$\begin{gathered} I=\frac{1}{\sqrt{3}} \int \frac{1}{\sqrt{t^{2}+\left(\frac{\sqrt{59}}{6}\right)^{2}}} d x \\ =\frac{1}{\sqrt{3}} \log \left|t+\sqrt{t^{2}+\left(\frac{\sqrt{59}}{6}\right)^{2}}\right|+c \\ {\left[\because \int \frac{1}{\sqrt{x^{2}+a^{2}}} d x=\log \left|x+\sqrt{x^{2}+a^{2}}\right|+c\right]} \end{gathered}$

\begin{aligned} &=\frac{1}{\sqrt{3}} \log \left(x+\frac{5}{6}\right)+\sqrt{\left(x+\frac{5}{6}\right)^{2}+\frac{59}{36}} \mid+c \\ &=\frac{1}{\sqrt{3}} \log \left(x+\frac{5}{6}\right)+\sqrt{(x)^{2}+\left(\frac{5}{6}\right)^{2}+2 \cdot \frac{5}{6} x+\frac{59}{36} \mid+c} \quad\quad\quad\quad\quad\quad\quad\left[\because t=x+\frac{5}{6}\right] \\ &=\frac{1}{\sqrt{3}} \log \left(x+\frac{5}{6}\right)+\sqrt{x^{2}+\frac{5 x}{3}+\frac{25}{36}+\frac{59}{36}} \mid+c \end{aligned}

\begin{aligned} &=\frac{1}{\sqrt{3}} \log \left(x+\frac{5}{6}\right)+\sqrt{x^{2}+\frac{5 x}{3}+\frac{84}{36}} \mid+c \\ &=\frac{1}{\sqrt{3}} \log \left(x+\frac{5}{6}\right)+\sqrt{x^{2}+\frac{5 x}{3}+\frac{7}{3}} \mid+c \end{aligned}