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EXAMPLE 6 Evaluate Solution We first separate the integrand to get [ egin{array}{ll} & int frac{3 x+2}{sqrt{1-x^{2}}} d x . \ int frac{3 x+2}{sqrt{1-x^{2}}} d x=3 int frac{x d x}{sqrt{1-x^{2}}}+2 int frac{d x}{sqrt{1-x^{2}}} . & end{array} ] In the first of these new integrals, we substitute [ u=1-x^{2}, quad d u=-2 x d x, quad ext { so } quad x d x=-frac{1}{2} d u . ] Then we obtain [ egin{aligned} 3 int frac{x d x}{sqrt{1-x^{2}}} & =3 int frac{(-1 / 2) d u}{sqrt{u}}=-frac{3}{2} int u^{-1 / 2} d u \ & =-frac{3}{2} cdot frac{u^{1 / 2}}{1 / 2}+C_{1}=-3 sqrt{1-x^{2}}+C_{1} . end{aligned} ] The second of the new integrals is a standard form, [ 2 int frac{d x}{sqrt{1-x^{2}}}=2 sin ^{-1} x+C_{2} . quad ext { Table 8.1, Formula } 18 ] Combining these results and renaming ( C_{1}+C_{2} ) as ( C ) gives [ int frac{3 x+2}{sqrt{1-x^{2}}} d x=-3 sqrt{1-x^{2}}+2 sin ^{-1} x+C . ]

          EXAMPLE 6 Evaluate

Solution We first separate the integrand to get
[
egin{array}{ll} 
& int frac{3 x+2}{sqrt{1-x^{2}}} d x . \
int frac{3 x+2}{sqrt{1-x^{2}}} d x=3 int frac{x d x}{sqrt{1-x^{2}}}+2 int frac{d x}{sqrt{1-x^{2}}} . &
end{array}
]

In the first of these new integrals, we substitute
[
u=1-x^{2}, quad d u=-2 x d x, quad 	ext { so } quad x d x=-frac{1}{2} d u .
]

Then we obtain
[
egin{aligned}
3 int frac{x d x}{sqrt{1-x^{2}}} & =3 int frac{(-1 / 2) d u}{sqrt{u}}=-frac{3}{2} int u^{-1 / 2} d u \
& =-frac{3}{2} cdot frac{u^{1 / 2}}{1 / 2}+C_{1}=-3 sqrt{1-x^{2}}+C_{1} .
end{aligned}
]

The second of the new integrals is a standard form,
[
2 int frac{d x}{sqrt{1-x^{2}}}=2 sin ^{-1} x+C_{2} . quad 	ext { Table 8.1, Formula } 18
]

Combining these results and renaming ( C_{1}+C_{2} ) as ( C ) gives
[
int frac{3 x+2}{sqrt{1-x^{2}}} d x=-3 sqrt{1-x^{2}}+2 sin ^{-1} x+C .
]

$EXAMPLE 6 Evaluate Solution We first separate the integrand to get [ eginarrayll int frac3 x+2sqrt1-x^2 d x . int frac3 x+2sqrt1-x^2 d x=3 int fracx d xsqrt1-x^2+2 int fracd xsqrt1-x^2 . endarray ] In the first of these new integrals, we substitute [ u=1-x^2, quad d u=-2 x d x, quad ext so quad x d x=-frac12 d u . ] Then we obtain [ eginaligned 3 int fracx d xsqrt1-x^2 =3 int frac(-1 / 2) d usqrtu=-frac32 int u^-1 / 2 d u =-frac32 cdot fracu^1 / 21 / 2+C1=-3 sqrt1-x^2+C1 . endaligned ] The second of the new integrals is a standard form, [ 2 int fracd xsqrt1-x^2=2 sin ^-1 x+C2 . quad ext Table 8.1, Formula 18 ] Combining these results and renaming ( C1+C2 ) as ( C ) gives [ int frac3 x+2sqrt1-x^2 d x=-3 sqrt1-x^2+2 sin ^-1 x+C . ]$

Added by Muhamad J.

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