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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 75 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 75 Maths Textbook Solution.

Answers (1)

Answer:

-3 \sqrt{6+x-2 x^{2}}+\frac{13}{2 \sqrt{2}} \sin ^{-1}\left(\frac{4 x-1}{7}\right)+c

Hint:

To solve the given solution we will use the partial fraction.

Given:

\int \frac{6 x+5}{\sqrt{6+x-2 x^{2}}}

Solution:

\int \frac{6 x+5}{\sqrt{6+x-2 x^{2}}}

   =\int \frac{\frac{-8}{2}(-4 x+1)+\left(\frac{13}{2}\right)}{\sqrt{6+x-2 x^{2}}} d x

   =-\frac{3}{2}\left[\int \frac{1-4 x}{\sqrt{6+x-2 x^{2}}} d x+\frac{13}{2} \int \frac{1}{\sqrt{6+x-2 x^{2}}} d x\right.

  =-\frac{3}{2}\left[\int \frac{d t}{\sqrt{t}}+\frac{13}{2 \sqrt{2}} \int \frac{1}{\sqrt{3+\left(\frac{x}{2}\right)-x^{2}}} d x\right.

=-\frac{3}{2} \times 2 \sqrt{t}+\frac{13}{2 \sqrt{2}} \int \frac{1}{\sqrt{3-\left(x^{2}-\left(\frac{x}{2}\right)+\frac{1}{16}-\frac{1}{16}\right)}}

=-3 \sqrt{6+x-2 x^{2}}+\frac{13}{2 \sqrt{2}} \int \frac{1}{\sqrt{3+\left(\frac{1}{16}\right)-\left(x-\left(\frac{1}{4}\right)\right)^{2}}}

=-3 \sqrt{6+x-2 x^{2}}+\frac{13}{2 \sqrt{2}} \int \frac{1}{\sqrt{\left(\frac{7}{4}\right)^{2}-\left(x-\left(\frac{1}{4}\right)\right)^{2}}} d x

=-3 \sqrt{6+x-2 x^{2}}+\frac{13}{2 \sqrt{2}} \sin ^{-1}\left(\frac{x-\left(\frac{1}{4}\right)}{\frac{7}{4}}\right)+c

=-3 \sqrt{6+x-2 x^{2}}+\frac{13}{2 \sqrt{2}} \sin ^{-1}\left(\frac{4 x-1}{7}\right)+c

 

Posted by

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