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  • Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 47 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise  Question 47 Maths Textbook Solution.

Answers (1)

Answer:

\frac{1}{17} \log \left|\frac{x-\frac{2}{8}}{x+5}\right|+c

Given:

\int \frac{1}{3 x^{2}+13 x-10} d x

Hint:

In this equation we have to make whole square statement from quadratic equation.

Solution:

I=\frac{1}{3} \int \frac{1}{x^{2}+\frac{18 x}{8}-\frac{10}{8}} d x

   I=\frac{1}{3} \int \frac{1}{\left(x+\frac{18}{6}\right)^{2}-\frac{169}{36}-\frac{10}{3}} d x

I=\frac{1}{3} \int \frac{d x}{\left(x+\frac{18}{6}\right)^{2}-\frac{169+120}{36}}

I=\frac{1}{3} \int \frac{d x}{\left(x+\frac{18}{6}\right)^{2}-\frac{289}{36}}

I=\frac{1}{3} \int \frac{d x}{\left(x+\frac{13}{6}\right)^{2}-\left(\frac{17}{6}\right)^{2}}

I=\frac{1}{3} \frac{1}{2 \times \frac{17}{6}} \log \left|\frac{x+\frac{18}{6}-\frac{17}{6}}{x+\frac{13}{6}+\frac{17}{6}}\right|+c

I=\frac{1}{17}+\log \left|\frac{x-\frac{4}{6}}{x+\frac{30}{6}}\right|+c

I=\frac{1}{17} \log \left|\frac{x-\frac{2}{8}}{x+5}\right|+c

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