Evaluate $int_1^efrac{1+x^2ln x}{x+x^2 ln x} dx$












7












$begingroup$



$$int_1^edfrac{1+x^2ln x}{x+x^2 ln x} dx$$




Attempt:



I have tried substitutions like $ln x = t$, but they are just not helping.



I end up with : $displaystyleint_0^1 dfrac{1+e^{2t}t}{1+ e^t t} dt $



Here method of substitution isn't really possible and integration by parts won't help. How else do I solve it?






calculus integration definite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    $ln x=t$ is a good substitution.
    $endgroup$
    – Nosrati
    Jul 24 '18 at 2:57
















7












$begingroup$



$$int_1^edfrac{1+x^2ln x}{x+x^2 ln x} dx$$




Attempt:



I have tried substitutions like $ln x = t$, but they are just not helping.



I end up with : $displaystyleint_0^1 dfrac{1+e^{2t}t}{1+ e^t t} dt $



Here method of substitution isn't really possible and integration by parts won't help. How else do I solve it?






calculus integration definite-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    $ln x=t$ is a good substitution.
    $endgroup$
    – Nosrati
    Jul 24 '18 at 2:57














7












7








7


4



$begingroup$



$$int_1^edfrac{1+x^2ln x}{x+x^2 ln x} dx$$




Attempt:



I have tried substitutions like $ln x = t$, but they are just not helping.



I end up with : $displaystyleint_0^1 dfrac{1+e^{2t}t}{1+ e^t t} dt $



Here method of substitution isn't really possible and integration by parts won't help. How else do I solve it?






calculus integration definite-integrals







share|cite|improve this question











$endgroup$





$$int_1^edfrac{1+x^2ln x}{x+x^2 ln x} dx$$




Attempt:



I have tried substitutions like $ln x = t$, but they are just not helping.



I end up with : $displaystyleint_0^1 dfrac{1+e^{2t}t}{1+ e^t t} dt $



Here method of substitution isn't really possible and integration by parts won't help. How else do I solve it?






calculus integration definite-integrals




calculus integration definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 13:49









Martin Sleziak

44.7k10119272




44.7k10119272










asked Jul 24 '18 at 2:53









AbcdAbcd

3,05331236




3,05331236












  • $begingroup$
    $ln x=t$ is a good substitution.
    $endgroup$
    – Nosrati
    Jul 24 '18 at 2:57


















  • $begingroup$
    $ln x=t$ is a good substitution.
    $endgroup$
    – Nosrati
    Jul 24 '18 at 2:57
















$begingroup$
$ln x=t$ is a good substitution.
$endgroup$
– Nosrati
Jul 24 '18 at 2:57




$begingroup$
$ln x=t$ is a good substitution.
$endgroup$
– Nosrati
Jul 24 '18 at 2:57










2 Answers
2






active

oldest

votes


















17












$begingroup$


The OP asked, "How else do I solve it?" We proceed to use a straightforward approach that circumvents the use of substitutions.





We need not use any substitutions. Rather, we can write



$$begin{align}
int_1^e frac{1+x^2log(x)}{x+x^2log(x)},dx&=int_1^e frac{1-x+x+x^2log(x)}{x+x^2log(x)},dx\\
&=(e-1)+int_1^e frac{1-x}{x+x^2log(x)},dx\\
&=(e-1)+int_1^e left(frac{1-x}{x+x^2log(x)}-frac1xright),dx+int_1^e frac1x,dx\\
&=e-int_1^e frac{1+log(x)}{1+xlog(x)},dx\\
&=e-left.left(log(1+xlog(x)) right)right|_1^e\\
&=e-log(1+e)
end{align}$$






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    Hi Mark ! May be, you could be more precise and recall that that $(1+xlog(x))'=1+log(x)$
    $endgroup$
    – Claude Leibovici
    Jul 24 '18 at 4:37



















4












$begingroup$

Let $$I = int frac{1+x^2ln x}{x+x^2ln x}dx$$



Divide both Numerator and Denominator by $x^2$



so $$I = int frac{frac{1}{x^2}+ln x}{frac{1}{x}+ln x}dx = int frac{bigg(frac{1}{x}+ln xbigg)-bigg(frac{1}{x}-frac{1}{x^2}bigg)}{frac{1}{x}+ln x}dx$$



so $$I = x-ln bigg|frac{1}{x}+ln xbigg|+mathcal{C}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    (+1) Nicely done.
    $endgroup$
    – Mark Viola
    Nov 5 '18 at 16:22











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









17












$begingroup$


The OP asked, "How else do I solve it?" We proceed to use a straightforward approach that circumvents the use of substitutions.





We need not use any substitutions. Rather, we can write



$$begin{align}
int_1^e frac{1+x^2log(x)}{x+x^2log(x)},dx&=int_1^e frac{1-x+x+x^2log(x)}{x+x^2log(x)},dx\\
&=(e-1)+int_1^e frac{1-x}{x+x^2log(x)},dx\\
&=(e-1)+int_1^e left(frac{1-x}{x+x^2log(x)}-frac1xright),dx+int_1^e frac1x,dx\\
&=e-int_1^e frac{1+log(x)}{1+xlog(x)},dx\\
&=e-left.left(log(1+xlog(x)) right)right|_1^e\\
&=e-log(1+e)
end{align}$$






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    Hi Mark ! May be, you could be more precise and recall that that $(1+xlog(x))'=1+log(x)$
    $endgroup$
    – Claude Leibovici
    Jul 24 '18 at 4:37
















17












$begingroup$


The OP asked, "How else do I solve it?" We proceed to use a straightforward approach that circumvents the use of substitutions.





We need not use any substitutions. Rather, we can write



$$begin{align}
int_1^e frac{1+x^2log(x)}{x+x^2log(x)},dx&=int_1^e frac{1-x+x+x^2log(x)}{x+x^2log(x)},dx\\
&=(e-1)+int_1^e frac{1-x}{x+x^2log(x)},dx\\
&=(e-1)+int_1^e left(frac{1-x}{x+x^2log(x)}-frac1xright),dx+int_1^e frac1x,dx\\
&=e-int_1^e frac{1+log(x)}{1+xlog(x)},dx\\
&=e-left.left(log(1+xlog(x)) right)right|_1^e\\
&=e-log(1+e)
end{align}$$






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    Hi Mark ! May be, you could be more precise and recall that that $(1+xlog(x))'=1+log(x)$
    $endgroup$
    – Claude Leibovici
    Jul 24 '18 at 4:37














17












17








17





$begingroup$


The OP asked, "How else do I solve it?" We proceed to use a straightforward approach that circumvents the use of substitutions.





We need not use any substitutions. Rather, we can write



$$begin{align}
int_1^e frac{1+x^2log(x)}{x+x^2log(x)},dx&=int_1^e frac{1-x+x+x^2log(x)}{x+x^2log(x)},dx\\
&=(e-1)+int_1^e frac{1-x}{x+x^2log(x)},dx\\
&=(e-1)+int_1^e left(frac{1-x}{x+x^2log(x)}-frac1xright),dx+int_1^e frac1x,dx\\
&=e-int_1^e frac{1+log(x)}{1+xlog(x)},dx\\
&=e-left.left(log(1+xlog(x)) right)right|_1^e\\
&=e-log(1+e)
end{align}$$






share|cite|improve this answer











$endgroup$




The OP asked, "How else do I solve it?" We proceed to use a straightforward approach that circumvents the use of substitutions.





We need not use any substitutions. Rather, we can write



$$begin{align}
int_1^e frac{1+x^2log(x)}{x+x^2log(x)},dx&=int_1^e frac{1-x+x+x^2log(x)}{x+x^2log(x)},dx\\
&=(e-1)+int_1^e frac{1-x}{x+x^2log(x)},dx\\
&=(e-1)+int_1^e left(frac{1-x}{x+x^2log(x)}-frac1xright),dx+int_1^e frac1x,dx\\
&=e-int_1^e frac{1+log(x)}{1+xlog(x)},dx\\
&=e-left.left(log(1+xlog(x)) right)right|_1^e\\
&=e-log(1+e)
end{align}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jul 24 '18 at 3:11

























answered Jul 24 '18 at 3:04









Mark ViolaMark Viola

133k1277174




133k1277174








  • 2




    $begingroup$
    Hi Mark ! May be, you could be more precise and recall that that $(1+xlog(x))'=1+log(x)$
    $endgroup$
    – Claude Leibovici
    Jul 24 '18 at 4:37














  • 2




    $begingroup$
    Hi Mark ! May be, you could be more precise and recall that that $(1+xlog(x))'=1+log(x)$
    $endgroup$
    – Claude Leibovici
    Jul 24 '18 at 4:37








2




2




$begingroup$
Hi Mark ! May be, you could be more precise and recall that that $(1+xlog(x))'=1+log(x)$
$endgroup$
– Claude Leibovici
Jul 24 '18 at 4:37




$begingroup$
Hi Mark ! May be, you could be more precise and recall that that $(1+xlog(x))'=1+log(x)$
$endgroup$
– Claude Leibovici
Jul 24 '18 at 4:37











4












$begingroup$

Let $$I = int frac{1+x^2ln x}{x+x^2ln x}dx$$



Divide both Numerator and Denominator by $x^2$



so $$I = int frac{frac{1}{x^2}+ln x}{frac{1}{x}+ln x}dx = int frac{bigg(frac{1}{x}+ln xbigg)-bigg(frac{1}{x}-frac{1}{x^2}bigg)}{frac{1}{x}+ln x}dx$$



so $$I = x-ln bigg|frac{1}{x}+ln xbigg|+mathcal{C}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    (+1) Nicely done.
    $endgroup$
    – Mark Viola
    Nov 5 '18 at 16:22
















4












$begingroup$

Let $$I = int frac{1+x^2ln x}{x+x^2ln x}dx$$



Divide both Numerator and Denominator by $x^2$



so $$I = int frac{frac{1}{x^2}+ln x}{frac{1}{x}+ln x}dx = int frac{bigg(frac{1}{x}+ln xbigg)-bigg(frac{1}{x}-frac{1}{x^2}bigg)}{frac{1}{x}+ln x}dx$$



so $$I = x-ln bigg|frac{1}{x}+ln xbigg|+mathcal{C}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    (+1) Nicely done.
    $endgroup$
    – Mark Viola
    Nov 5 '18 at 16:22














4












4








4





$begingroup$

Let $$I = int frac{1+x^2ln x}{x+x^2ln x}dx$$



Divide both Numerator and Denominator by $x^2$



so $$I = int frac{frac{1}{x^2}+ln x}{frac{1}{x}+ln x}dx = int frac{bigg(frac{1}{x}+ln xbigg)-bigg(frac{1}{x}-frac{1}{x^2}bigg)}{frac{1}{x}+ln x}dx$$



so $$I = x-ln bigg|frac{1}{x}+ln xbigg|+mathcal{C}.$$






share|cite|improve this answer









$endgroup$



Let $$I = int frac{1+x^2ln x}{x+x^2ln x}dx$$



Divide both Numerator and Denominator by $x^2$



so $$I = int frac{frac{1}{x^2}+ln x}{frac{1}{x}+ln x}dx = int frac{bigg(frac{1}{x}+ln xbigg)-bigg(frac{1}{x}-frac{1}{x^2}bigg)}{frac{1}{x}+ln x}dx$$



so $$I = x-ln bigg|frac{1}{x}+ln xbigg|+mathcal{C}.$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jul 24 '18 at 14:22









DXTDXT

5,9022731




5,9022731












  • $begingroup$
    (+1) Nicely done.
    $endgroup$
    – Mark Viola
    Nov 5 '18 at 16:22


















  • $begingroup$
    (+1) Nicely done.
    $endgroup$
    – Mark Viola
    Nov 5 '18 at 16:22
















$begingroup$
(+1) Nicely done.
$endgroup$
– Mark Viola
Nov 5 '18 at 16:22




$begingroup$
(+1) Nicely done.
$endgroup$
– Mark Viola
Nov 5 '18 at 16:22


















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