Question

Evaluate $\int 2^{x+1} -5^{x-1} /10^{x} dx$

Answer 1

$\large{\mathfrak{\underline{\underline{Question:-}}}}$

$\bigstar\int \frac { {2} ^{x+1} - {5} ^{x-1} }{ {10} ^{x} } dx$

$\large{\mathfrak{\underline{\underline{Answer:-}}}}$

$= \frac{1}{5 log2 \: . \: {2}^{x} } - \frac{1}{2 . log5 \: . \: {5}^{x} } + c$

$\large{\mathfrak{\underline{\underline{Explanation:-}}}}$

$\int \frac { {2} ^{x+1} - {5} ^{x-1} }{ {10} ^{x} } dx$

$= \int \frac{ {2}^{x }\: . \: {2}^{1} - {5}^{x} \: . \: {5}^{ - 1} }{ {2}^{x} \: . \: {5}^{x} } dx$

$= \int \frac{2 \: . \: {2}^{x} }{ {2}^{x} \: . \: {5}^{x} } - \frac{ {5}^{ -1} \: . \: {5}^{x} }{ {2}^{x} \: . \: {5}^{x} } dx$

$= 2 \int {5}^{ - x} dx - \frac{1}{5} \int {2}^{ - x} dx$

$= - 2 \: . \: \frac{ {5}^{x} }{ loge5 } + \frac{1}{5} \: . \: \frac{ {2}^{ - x} }{log2} + c$

$= \frac{1}{5 log 2\: . \: {2}^{x} } - \frac{1}{2 \: . \: log5 \: . \: {5}^{x} } + c$

$\large{\mathfrak{\underline{\underline{Varification:-}}}}$

$\implies\int \frac { {2} ^{x+1} - {5} ^{x-1} }{ {10} ^{x} } dx$

$\implies\frac{1}{5 log 2\: . \: {2}^{x} } - \frac{1}{2 \: . \: log5 \: . \: {5}^{x} } + c$

Hence proved!