Question

Integrate:-

$\displaystyle\sf\int\dfrac{1}{x^6+1}\:dx$​

Answer 1

SOLUTION

We have,

$\large{\displaystyle{\sf l \: = \int \dfrac{x}{\sqrt[6]{x + 1}}dx}}l=∫$

The above integral can be solved by using u substitution

Now,

Assume,

$\tt\: \: u = x + 1u=x+1$

Taking derivative both sides,we get :

$\leadsto \: \boxed{ \boxed{\bf \: du = dx} }$

The integral can be rewritten as :

$\begin{gathered}\displaystyle{\sf l \: = \int \dfrac{u - 1}{\sqrt[6]{u}} \: du}\\ \\ \longrightarrow \: \displaystyle{ \sf \: l = \int \: (u - 1) {u}^{ \frac{1}{6} }.du } \\ \\ \longrightarrow \: \displaystyle{ \sf \: l = \int \: (u {}^{ \frac{7}{6} - } u {}^{ \frac{1}{6} }).du }\end{gathered} </p><p>$

Applying the formula of integrate :

$\begin{gathered} \longrightarrow \: \displaystyle{ \sf l = \: \dfrac{u {}^{ \frac{7}{6} + 1 }}{ \frac{7}{6} + 1} - \dfrac{ {u}^{ \frac{1}{6} + 1} }{ \frac{1}{6} + 1 } + c } \\ \\ \longrightarrow \: \sf \: l = \dfrac{6 {u}^{ \frac{13}{6} } }{13} - \dfrac{6 {u}^{ \frac{7}{6} } }{7} + c \\ \\ \longrightarrow \: \boxed{ \boxed{\sf \: l = \frac{6( \sqrt[6]{x + 1} ) {}^{13} }{7} - \frac{6( \sqrt[6]{x + 1}) {}^{7} }{7} + c}}\end{gathered}$

Answer 2

Answer

AnswerOpen in answr app

AnswerOpen in answr app∫x(x6+1)1dx

AnswerOpen in answr app∫x(x6+1)1dx=∫x(x6+1)x6+1−x6dx

AnswerOpen in answr app∫x(x6+1)1dx=∫x(x6+1)x6+1−x6dx=∫x(x6+1)x6+1dx−∫x(x6+1)x6dx

AnswerOpen in answr app∫x(x6+1)1dx=∫x(x6+1)x6+1−x6dx=∫x(x6+1)x6+1dx−∫x(x6+1)x6dx=∫xdx−∫(x6+1)x5dx

AnswerOpen in answr app∫x(x6+1)1dx=∫x(x6+1)x6+1−x6dx=∫x(x6+1)x6+1dx−∫x(x6+1)x6dx=∫xdx−∫(x6+1)x5dxLet t=x6+1⇒dt=6x5dx

AnswerOpen in answr app∫x(x6+1)1dx=∫x(x6+1)x6+1−x6dx=∫x(x6+1)x6+1dx−∫x(x6+1)x6dx=∫xdx−∫(x6+1)x5dxLet t=x6+1⇒dt=6x5dx=∫xdx−61∫tdt

AnswerOpen in answr app∫x(x6+1)1dx=∫x(x6+1)x6+1−x6dx=∫x(x6+1)x6+1dx−∫x(x6+1)x6dx=∫xdx−∫(x6+1)x5dxLet t=x6+1⇒dt=6x5dx=∫xdx−61∫tdt=logx−61log∣t∣+c

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