Question

integration to (a^x+b^x)^2/a^x b^x​

Answer 1

$\textbf{Given:}$

$\displaystyle\int\,\frac{(a^x+b^x)^2}{a^xb^x}\;dx$

$=\displaystyle\int\,\frac{(a^{2x}+b^{2x}+2\,a^xb^x)}{a^xb^x}\;dx$

$=\displaystyle\int\,[\frac{a^{2x}}{a^xb^x}+\frac{b^{2x}}{a^xb^x}+\frac{2\a^xb^x}{a^xb^x}]\;dx$

$=\displaystyle\int\,[\frac{a^x}{b^x}+\frac{b^x}{a^x}+2]\;dx$

$=\displaystyle\int\,[(\frac{a}{b})^x+(\frac{b}{a})^x+2]\;dx$

$=\displaystyle\int\,(\frac{a}{b})^x\,dx+\int\,(\frac{b}{a})^x\,dx+2\int\,dx$

$\text{We know that,}$

$\boxed{\bf\int\,a^x\,dx=\frac{a^x}{loga}+c}$

$=\displaystyle\,\frac{(\frac{a}{b})^x}{log\frac{a}{b}}+\frac{(\frac{b}{a})^x}{log\frac{b}{a}}+2x+c$

$\therefore\bf\,\displaystyle\int\,\frac{(a^x+b^x)^2}{a^xb^x}\;dx=\frac{(\frac{a}{b})^x}{log(\frac{a}{b})}+\frac{(\frac{b}{a})^x}{log(\frac{b}{a})}+2x+c$

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