Question

HLP pls... !!

Answer 1

\[\int\limits\limits_{}^{}(ax ^{2} - b)dx/[x \sqrt{c ^{2}x ^{2}- (ax ^{2}+ b)^{2}}]\]

Answer 2

also if you want to write fractions in equation, use \frac{x}{y}
to get
\[\frac{x}{y}\]

Answer 3

ya d bracket part is the denominator @completeidiot

Answer 4

\[\int\limits_{}^{}\frac{ax^2-b}{x\sqrt{c^2x^2-(ax^2+b)^2}}dx\]

Answer 5

yep .. !!

Answer 6

what are the direction

Answer 7

north

Answer 8

integration is like finding the right key for the lock, finding the right substitution :)

Answer 9

\[Let I =\int\limits \frac{\left( ax ^{2} -b\right) }{x \sqrt{c ^{2}x ^{2}-\left( ax ^{2}+b \right)^{2}} }dx\]
\[=\int\limits \frac{\left( ax ^{2}-b \right) }{x ^{2}\sqrt{c ^{2}-\left( \frac{ ax ^{2}+b }{ x } \right)^{2}} }dx\]
\[=\int\limits \frac{ \left( ax ^{2}-b \right) }{ x ^{2}\sqrt{c ^{2}-\left( ax+\frac{ b }{ x } \right)^{2}} }dx\]
\[plug ax+\frac{ b }{x }=c \sin \theta \]
diff.
\[\left( a-\frac{ b }{x ^{2} } \right)dx=c \cos \theta d \theta \]
\[\frac{ \left( ax ^{2}-b \right) }{x ^{2} }dx=c \cos \theta d \theta \]
\[Therefore I=\int\limits \frac{ c \cos \theta d \theta }{\sqrt{c ^{2}-c ^{2}\sin ^{2}\theta } }\]
\[I=\int\limits \frac{ c \cos \theta d \theta }{c \cos \theta }=\theta +c1\]
\[I=\sin^{-1} \frac{ ax ^{2}+b }{cx}+c1\]