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  • Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.21 Question 3 maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.21 Question 3 maths Textbook Solution.

Answers (1)

Answer:-\sqrt{4+5 x-x^{2}}+\frac{7}{2} \sin ^{-1}\left(\frac{2 x-5}{\sqrt{41}}\right)+c

Given: \int \frac{x+1}{\sqrt{-x^{2}+5 x+4}} d x

Hint: Simplify it and solve it

Solution: I=\int \frac{x+1}{\sqrt{-x^{2}+5 x+4}} d x

               I=\int \frac{x+1}{\sqrt{4-\left(x^{2}-5 x\right)}} d x=\int \frac{x+1}{\sqrt{4+\left(\frac{5}{2}\right)^{2}-\left(x^{2}-5 x+\left(\left(\frac{5}{2}\right)^{2}\right)\right.}} d x

             I=\int \frac{x+1}{\sqrt{\frac{41}{4}-\left(x-\frac{5}{2}\right)^{2}}} d x

            I=\frac{1}{2} \int \frac{2 x-5+7}{\sqrt{\frac{41}{4}-\left(x-\frac{5}{2}\right)^{2}}} d x

            =\frac{1}{2} \int \frac{2 x-5}{\sqrt{\frac{41}{4}-\left(x-\frac{5}{2}\right)^{2}}} d x+\frac{7}{2} \int \frac{d x}{\sqrt{\frac{41}{4}-\left(x-\frac{5}{2}\right)^{2}}}

            \frac{41}{4}-\left(x-\frac{5}{2}\right)^{2}=y

           \begin{aligned} &-2\left(x-\frac{5}{2}\right) d x=d y \\ &(2 x-5) d x=-d y \end{aligned}

            \Rightarrow I=\frac{1}{2} \int \frac{-d y}{\sqrt{y}}+\frac{7}{2} \int \frac{d x}{\sqrt{\frac{41}{4}-\left(x-\frac{5}{2}\right)^{2}}}

            \Rightarrow I=-\frac{1}{2}\left[\frac{(y)^{\frac{1}{2}}}{\frac{1}{2}}\right]+\frac{7}{2}\left(\sin ^{-1}\left(\frac{x-\frac{5}{2}}{\frac{\sqrt{41}}{2}}\right)\right)+c \quad\left[\int \frac{1}{\sqrt{a^{2}-x^{2}}}=\sin ^{-1}\left(\frac{x}{a}\right)+c\right]

           \begin{aligned} &\Rightarrow I=-\left[\frac{41}{4}-\left(x-\frac{5}{2}\right)^{2}\right]^{\frac{1}{2}}+\frac{7}{2} \sin ^{-1}\left(\frac{2 x-5}{\sqrt{41}}\right)+c \\ &\Rightarrow I=-\sqrt{5 x+4-x^{2}}+\frac{7}{2} \sin ^{-1}\left(\frac{2 x-5}{\sqrt{41}}\right)+c \end{aligned}

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