Integrals

Hello, bigton!

There is a formula for these exponential integrals:

. . $\int b^u\,du \;=\;\frac{b^u}{\ln(b)}+ C$

Quote:

$1)\;\;\int^3_{\text{-}2}(25)^{2x+1} dx$

Let: $u \:=\:2x+1\quad\Rightarrow\quad du \:=\:2\,dx \quad\Rightarrow\quad dx \:=\:\tfrac{1}{2}du$

Substitute: . $\int25^u\cdot\tfrac{1}{2}du \;=\;\tfrac{1}{2}\int 25^u\,du \;=\;\frac{1}{2}\,\frac{25^u}{\ln(25)} + C$

Back-substitute: . $\frac{25^{2x+1}}{2\ln(25)}\,\bigg]^3_{\text{-}2} \;=\;\frac{25^7}{2\ln(25)} - \frac{25^{-1}}{2\ln(25)} \quad\hdots\;\text{ etc.}$

Quote:

$2)\;\;\int 7x\!\cdot\!6^{1-2x^2}\,dx$

We have: . $7\!\int 6^{1-2x^2}(x\,dx)$

Let: $u \:=\:1-2x^2\quad\Rightarrow\quad du \:=\:\text{-}4x\,dx \quad\Rightarrow\quad x\,dx \:=\:\text{-}\tfrac{1}{4}du$

Substitute: . $7\!\int 6^u(\text{-}\tfrac{1}{4}du) \;=\;-\tfrac{7}{4}\!\int6^u\,du \;=\;-\frac{7}{4}\,\frac{6^u}{\ln(6)} + C$

Back-substitute: . $-\frac{7\cdot6^{1-2x^2}}{4\cdot\ln(6)} + C$