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  • Provide Solution for RD Sharma Class 12 Chapter Indefinite Integrals Exercise 18.28 Question 10

Provide Solution for RD Sharma Class 12 Chapter Indefinite Integrals Exercise 18.28 Question 10

Answers (1)

best_answer

Answer:-

\frac{4 x+3}{8} \sqrt{2 x^{2}+3 x+4}+\frac{23 \sqrt{2}}{32} \log \left|x+\frac{3}{4} \sqrt{x^{2}-\frac{3}{2}+2}\right|+c

Hint:-

Multiplying by  2\sqrt{2}

Given:-

\int \sqrt{2 x^{2}+3 x+4} d x

Solution:-

\begin{aligned} &=\frac{1}{2 \sqrt{2}} \int \sqrt{16 x^{2}+24 x+32} d x \\\\ &=\frac{1}{2 \sqrt{2}} \int \sqrt{(4 x)^{2}+2.4 x \cdot 3+(3)^{2}+23} d x \\\\ &=\frac{1}{2 \sqrt{2}} \int \sqrt{(4 x+3)^{2}+\sqrt{23}^{2}} d x \end{aligned}

 

By using the formula

 

\begin{aligned} &\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}+a^{2}}\right|+c \\\\ &=\frac{1}{2 \sqrt{2}}\left[\frac{1}{2} \times \frac{1}{4} \times(4 x+3) \sqrt{(4 x+3)^{2}}+\frac{\sqrt{23}^{2}}{2} \times \frac{1}{4} \log \left[(4 x+3)+\sqrt{(4 x+3)^{2}+\sqrt{23}}\right]+c\right] \end{aligned}

\begin{aligned} &=\frac{4 x+3}{8} \sqrt{2 x^{2}+3 x+4}+\frac{23 \sqrt{2}}{32} \log \left|(4 x+3)+\sqrt{2 x^{2}+3 x+4}\right|+c \\\\ &=\frac{4 x+3}{8} \sqrt{2 x^{2}+3 x+4}+\frac{23 \sqrt{2}}{32} \log \left|\left(x+\frac{3}{4}\right) \sqrt{x^{2}-\frac{3}{2}+2}\right|+\frac{23 \sqrt{2}}{32} \log 4+c \\\\ &=\frac{4 x+3}{8} \sqrt{2 x^{2}+3 x+4}+\frac{23 \sqrt{2}}{32} \log \left|x+\frac{3}{4} \sqrt{x^{2}-\frac{3}{2}+2}\right|+c \end{aligned}

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