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Find the integral \int \left ( x+4 \right )\sqrt{x^{2}+8x-9}dx

  • Option 1)

    \frac{1}{3}\left ( x^{2}+8x-9 \right )^{\frac{3}{2}}+C

  • Option 2)

    \frac{1}{2}\left ( x^{2}+8x-9 \right )^{\frac{3}{2}}+C

  • Option 3)

    \frac{1}{3}\left ( x^{2}+8x-9 \right )^{\frac{5}{2}}+C

  • Option 4)

    \frac{1}{6}\left ( x^{2}+8x-9 \right )^{\frac{3}{2}}+C

 

Answers (1)

best_answer

As we learned,

 

Type of Integration by perfect square -

The integral in the form of :

(i)    \int \frac{px+r}{ax^{2}+bx+c}dx     (ii) \int \frac{px+r}{\sqrt{{ax^{2}+bx+c}}}dx

(iii) \int (px+r){\sqrt{{ax^{2}+bx+c}}}dx  

\therefore \int \frac{px+r}{ax^{2}+bx+c}dx=A\frac{\ d.c.\ of\left ( ax^{2}+bx+c \right )}{ax^{2}+bx+c}+B\cdot \frac{1}{ax^{2}+bx+c}  and find integrals by standard formulae

 

- wherein

Working rule.

Let     px+r=A\frac{\mathrm{d} }{\mathrm{d} x}(ax^{2}+bx+c)+B

Find A and B by comparing

 

 

 

Put  x^{2}+8x-9=t

2\left ( x+4 \right )dx=dt

\int \frac{1}{2}\sqrt{t}dt=\frac{1}{3}t^{\frac{3}{2}}+C


Option 1)

\frac{1}{3}\left ( x^{2}+8x-9 \right )^{\frac{3}{2}}+C

Option 2)

\frac{1}{2}\left ( x^{2}+8x-9 \right )^{\frac{3}{2}}+C

Option 3)

\frac{1}{3}\left ( x^{2}+8x-9 \right )^{\frac{5}{2}}+C

Option 4)

\frac{1}{6}\left ( x^{2}+8x-9 \right )^{\frac{3}{2}}+C

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