Question

ANSWER this please...​

Answer 1

$=\displaystyle\int\,\frac{(a^{2x}+b^{2x}+2\,a^xb^x)}{a^xb^x}\;dx\\\\=\displaystyle\int\,[\frac{ a^{2x}}{ a^xb^x}+\frac{b^{2x}}{ a^xb^x}+\frac{2 a^xb^x}{a^xb^x}]\;dx \\\\=\displaystyle\int\,[\frac{a^x}{b^x}+\frac{b^x}{a^x}]\\\\=\displaystyle\int\,[(\frac{a}{b})^x+(\frac{b}{a})]\\\\=\displaystyle\int\,(\frac{a}{b})^x\,dx+\int\,(\frac{b}{a})^x\,dx+2\int\,dx \\\\\sf{We \: know \: that,}\boxed{\bf\int\,a^x\,dx=\frac{a^x}{loga}+c}\\\\=\displaystyle\,\frac{(\frac{a}{b})^x}{log\frac{a}{b}}+\frac{(\frac{b}{a})^x}{log\frac{b}{a}}\\\\\therefore\bf\,\displaystyle\int\,\frac{(a^x+b^x)^2}{a^xb^x}\;dx=\frac{(\frac{a}{b})^x}{log(\frac{a}{b})}+\frac{(\frac{b}{a})^x}{log(\frac{b}{a})}$

Answer 2

=∫

a

x

b

x

(a

2x

+b

2x

+2a

x

b

x

)

dx

=∫[

a

x

b

x

a

2x

+

a

x

b

x

b

2x

+

a

x

b

x

2a

x

b

x

]dx

=∫[

b

x

a

x

+

a

x

b

x

]

=∫[(

b

a

)

x

+(

a

b

)]

=∫(

b

a

)

x

dx+∫(

a

)

x

dx+2∫dx

Weknowthat,

∫a

x

dx=

loga

a

x

+c

=

log

b

a

(

b

a

)

x

+

log

a

b

(

a

b

)

x

∴∫

a

x

b

x

(a

x

+b

x

)

2

dx=

log(

b

a

)

(

b

a

)

x

+

log(

a

b

)

(

a

b

)

x