Question

solve these Integration.​

Answer 1

Answer:

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Answer 2

Topic:

Integration

Formula used:

$\int\tan(x)\ dx = \log(\sec x) + C$

$\int\dfrac1x\ dx = \log(x) + C$

Solution 1:

We need to evaluate the following integral

$\displaystyle\int\dfrac{\tan(\ell n \ x)}{x}\, dx$

Since the derivative of ln(x) is 1/x, we can solve it using the method of substitution easily.

Put ln(x) = t so that dx/x = dt

The above integral changes to,

$\displaystyle\longrightarrow\int \tan(t)\ dt$

$\displaystyle \longrightarrow \ell n |\sec t| + C$

Undoing our substitution for t = ln(x), we get:
$\displaystyle \longrightarrow \ell n |\sec \ell n (x)| + C$

This is the required answer.

$\dotfill$

Solution 2:

We need to evaluate the following integral,

$\displaystyle \int \dfrac{1}{e^x +1}\ dx$

Put x = -u so that dx = du.

The above integral changes to,

$\displaystyle \longrightarrow\int \dfrac{-1}{e^{-u} +1}\ du$

$\displaystyle \longrightarrow\int \dfrac{-1}{\frac1{e^u} +1}\ du$

$\displaystyle \longrightarrow\int \dfrac{-1}{\frac{1 + e^u}{e^u}}\ du$

$\displaystyle \longrightarrow\int \dfrac{-e^u}{1 + e^u}\ du$

Again, put 1 + eᵘ = y so that eᵘ du = dy.

$\displaystyle \longrightarrow\int \dfrac{-dy}{y}$

$\displaystyle \longrightarrow - \log|y| + C$

Undoing our substitution,

$\displaystyle \longrightarrow - \log|1 + e^u| + C$

$\displaystyle \longrightarrow - \log|1 + e^{-x}| + C$

$\displaystyle \longrightarrow - \log\left|1 + \dfrac1{e^{x}}\right| + C$

$\displaystyle \longrightarrow - \log\left|\dfrac{e^x+1}{e^{x}}\right| + C$

$\displaystyle \longrightarrow \log\left|\dfrac{e^x}{e^{x} + 1}\right| + C$

$\displaystyle \longrightarrow \log|e^x| - \log|e^x + 1| + C$

$\displaystyle \longrightarrow x - \log|e^x + 1| + C$

This is the required answer.