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

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}