Question

Evaluate $\int \frac{(a^x-b^x)}{c^x}dx$

• A. $\frac{(a^x-b^x)}{c^x.log\left(ab\right)}+c$
• B. $\frac{1}{c^x}[\frac{a^x}{loga}-\frac{b^x}{logb}]+C$
• C. $\frac{1}{c^x}.(\frac{a^x}{loga}+\frac{b^x}{logb})+C$
• D. $\frac{1}{c^x}[\frac{a^x}{log\left(a/c\right)}-\frac{b^x}{log\left(b/c\right)}]+C$

Answer 1

Answer: (D)

$\frac{1}{c^x}[\frac{a^x}{log\left(a/c\right)}-\frac{b^x}{log\left(b/c\right)}]+C$

$I=\int \frac{a^x-b^x}{c^x}dx$

$=\int \left\{\right(\frac{a}{c}{)}^{2c}-\left(\frac{b}{c}{)}^x\right\}dx$

$=\int (\frac{a}{c}{)}^{x}dx-\int (\frac{b}{c}{)}^{x}dx$

$=I_1-I_2$

$I_1=\int (\frac{a}{c}{)}^{x}dx$

Let $x=\log (\frac{a}{c}{)}^t$

$dx=\frac{1}{t{\log }_e\left(a/c\right)}dt$ $[\because {\log }_{a}x=\frac{1}{{\log }_{e}a}]$

$I_1=\int (a/c{)}^{\log (a/c{)}^t}.\frac{1}{t{\log }_e\left(a/c\right)}dt=\int t.\frac{1}{t{\log }_e\left(a/c\right)}dt.$

$=\frac{1}{{\log }_e\left(a/c\right)}.\int dt$

$=\frac{t}{{\log }_e\left(a/c\right)}$

$=\frac{(a/c{)}^x}{{\log }_e\left(a/c\right)}$

Similarly,

$I_2=\frac{(b/c{)}^2}{{\log }_e\left(b/c\right)}$

$I=\frac{(a/c{)}^x}{{\log }_e\left(a/c\right)}+\frac{(b/c{)}^x}{{\log }_e\left(b/c\right)}+c$

$=\frac{1}{c^x}[\frac{a^x}{\log \left(a/c\right)}+\frac{b^x}{\log \left(b/c\right)}]+c$