Evaluate $int_1^edfrac1+x^2ln xx+x^2 ln x dx$

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$$int_1^edfrac1+x^2ln xx+x^2 ln x dx$$

Attempt:

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

I end up with : $displaystyleint_0^1 dfrac1+e^2tt1+ 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 definite-integrals

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edited Jul 24 at 12:33

asked Jul 24 at 2:53

Abcd

2,3361624

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $ln x=t$ is a good substitution.
  – Nosrati
  Jul 24 at 2:57
  

  
  
  
  

add a comment | 

up vote
5
down vote

favorite

3

$$int_1^edfrac1+x^2ln xx+x^2 ln x dx$$

Attempt:

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

I end up with : $displaystyleint_0^1 dfrac1+e^2tt1+ 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 definite-integrals

share|cite|improve this question

edited Jul 24 at 12:33

asked Jul 24 at 2:53

Abcd

2,3361624

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $ln x=t$ is a good substitution.
  – Nosrati
  Jul 24 at 2:57
  

  
  
  
  

add a comment | 

up vote
5
down vote

favorite

3

up vote
5
down vote

favorite

3

3

$$int_1^edfrac1+x^2ln xx+x^2 ln x dx$$

Attempt:

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

I end up with : $displaystyleint_0^1 dfrac1+e^2tt1+ 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 definite-integrals

share|cite|improve this question

edited Jul 24 at 12:33

asked Jul 24 at 2:53

Abcd

2,3361624

$$int_1^edfrac1+x^2ln xx+x^2 ln x dx$$

Attempt:

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

I end up with : $displaystyleint_0^1 dfrac1+e^2tt1+ 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 definite-integrals

share|cite|improve this question

edited Jul 24 at 12:33

asked Jul 24 at 2:53

Abcd

2,3361624

share|cite|improve this question

share|cite|improve this question

edited Jul 24 at 12:33

edited Jul 24 at 12:33

edited Jul 24 at 12:33

asked Jul 24 at 2:53

Abcd

2,3361624

asked Jul 24 at 2:53

Abcd

2,3361624

asked Jul 24 at 2:53

Abcd

2,3361624

2,3361624

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $ln x=t$ is a good substitution.
  – Nosrati
  Jul 24 at 2:57
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $ln x=t$ is a good substitution.
  – Nosrati
  Jul 24 at 2:57
  

  
  
  
  

$ln x=t$ is a good substitution.
– Nosrati
Jul 24 at 2:57

$ln x=t$ is a good substitution.
– Nosrati
Jul 24 at 2:57

add a comment | 

2 Answers
2

active

oldest

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up vote
16
down vote



accepted

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

$$beginalign
int_1^e frac1+x^2log(x)x+x^2log(x),dx&=int_1^e frac1-x+x+x^2log(x)x+x^2log(x),dx\\
&=(e-1)+int_1^e frac1-xx+x^2log(x),dx\\
&=(e-1)+int_1^e left(frac1-xx+x^2log(x)-frac1xright),dx+int_1^e frac1x,dx\\
&=e-int_1^e frac1+log(x)1+xlog(x),dx\\
&=e-left.left(log(1+xlog(x)) right)right|_1^e\\
&=e-log(1+e)
endalign$$

share|cite|improve this answer

edited Jul 24 at 3:11

answered Jul 24 at 3:04

Mark Viola

126k1172167

• 
  
  
  

  
  2
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

up vote
1
down vote

Let $$I = int frac1+x^2ln xx+x^2ln xdx$$

Divide both Numerator and Denominator by $x^2$

so $$I = int fracfrac1x^2+ln xfrac1x+ln xdx = int fracbigg(frac1x+ln xbigg)-bigg(frac1x-frac1x^2bigg)frac1x+ln xdx$$

so $$I = x-ln bigg|frac1x+ln xbigg|+mathcalC.$$

share|cite|improve this answer

answered Jul 24 at 14:22

Durgesh Tiwari

4,7862426

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
16
down vote



accepted

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

$$beginalign
int_1^e frac1+x^2log(x)x+x^2log(x),dx&=int_1^e frac1-x+x+x^2log(x)x+x^2log(x),dx\\
&=(e-1)+int_1^e frac1-xx+x^2log(x),dx\\
&=(e-1)+int_1^e left(frac1-xx+x^2log(x)-frac1xright),dx+int_1^e frac1x,dx\\
&=e-int_1^e frac1+log(x)1+xlog(x),dx\\
&=e-left.left(log(1+xlog(x)) right)right|_1^e\\
&=e-log(1+e)
endalign$$

share|cite|improve this answer

edited Jul 24 at 3:11

answered Jul 24 at 3:04

Mark Viola

126k1172167

• 
  
  
  

  
  2
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

up vote
16
down vote



accepted

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

$$beginalign
int_1^e frac1+x^2log(x)x+x^2log(x),dx&=int_1^e frac1-x+x+x^2log(x)x+x^2log(x),dx\\
&=(e-1)+int_1^e frac1-xx+x^2log(x),dx\\
&=(e-1)+int_1^e left(frac1-xx+x^2log(x)-frac1xright),dx+int_1^e frac1x,dx\\
&=e-int_1^e frac1+log(x)1+xlog(x),dx\\
&=e-left.left(log(1+xlog(x)) right)right|_1^e\\
&=e-log(1+e)
endalign$$

share|cite|improve this answer

edited Jul 24 at 3:11

answered Jul 24 at 3:04

Mark Viola

126k1172167

• 
  
  
  

  
  2
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

up vote
16
down vote



accepted

up vote
16
down vote



accepted

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

$$beginalign
int_1^e frac1+x^2log(x)x+x^2log(x),dx&=int_1^e frac1-x+x+x^2log(x)x+x^2log(x),dx\\
&=(e-1)+int_1^e frac1-xx+x^2log(x),dx\\
&=(e-1)+int_1^e left(frac1-xx+x^2log(x)-frac1xright),dx+int_1^e frac1x,dx\\
&=e-int_1^e frac1+log(x)1+xlog(x),dx\\
&=e-left.left(log(1+xlog(x)) right)right|_1^e\\
&=e-log(1+e)
endalign$$

share|cite|improve this answer

edited Jul 24 at 3:11

answered Jul 24 at 3:04

Mark Viola

126k1172167

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

$$beginalign
int_1^e frac1+x^2log(x)x+x^2log(x),dx&=int_1^e frac1-x+x+x^2log(x)x+x^2log(x),dx\\
&=(e-1)+int_1^e frac1-xx+x^2log(x),dx\\
&=(e-1)+int_1^e left(frac1-xx+x^2log(x)-frac1xright),dx+int_1^e frac1x,dx\\
&=e-int_1^e frac1+log(x)1+xlog(x),dx\\
&=e-left.left(log(1+xlog(x)) right)right|_1^e\\
&=e-log(1+e)
endalign$$

share|cite|improve this answer

edited Jul 24 at 3:11

answered Jul 24 at 3:04

Mark Viola

126k1172167

share|cite|improve this answer

share|cite|improve this answer

edited Jul 24 at 3:11

edited Jul 24 at 3:11

edited Jul 24 at 3:11

answered Jul 24 at 3:04

Mark Viola

126k1172167

answered Jul 24 at 3:04

Mark Viola

126k1172167

answered Jul 24 at 3:04

Mark Viola

126k1172167

126k1172167

• 
  
  
  

  
  2
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

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

  
  
  
  

2

2

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

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

add a comment | 

up vote
1
down vote

Let $$I = int frac1+x^2ln xx+x^2ln xdx$$

Divide both Numerator and Denominator by $x^2$

so $$I = int fracfrac1x^2+ln xfrac1x+ln xdx = int fracbigg(frac1x+ln xbigg)-bigg(frac1x-frac1x^2bigg)frac1x+ln xdx$$

so $$I = x-ln bigg|frac1x+ln xbigg|+mathcalC.$$

share|cite|improve this answer

answered Jul 24 at 14:22

Durgesh Tiwari

4,7862426

add a comment | 

up vote
1
down vote

Let $$I = int frac1+x^2ln xx+x^2ln xdx$$

Divide both Numerator and Denominator by $x^2$

so $$I = int fracfrac1x^2+ln xfrac1x+ln xdx = int fracbigg(frac1x+ln xbigg)-bigg(frac1x-frac1x^2bigg)frac1x+ln xdx$$

so $$I = x-ln bigg|frac1x+ln xbigg|+mathcalC.$$

share|cite|improve this answer

answered Jul 24 at 14:22

Durgesh Tiwari

4,7862426

add a comment | 

up vote
1
down vote

up vote
1
down vote

Let $$I = int frac1+x^2ln xx+x^2ln xdx$$

Divide both Numerator and Denominator by $x^2$

so $$I = int fracfrac1x^2+ln xfrac1x+ln xdx = int fracbigg(frac1x+ln xbigg)-bigg(frac1x-frac1x^2bigg)frac1x+ln xdx$$

so $$I = x-ln bigg|frac1x+ln xbigg|+mathcalC.$$

share|cite|improve this answer

answered Jul 24 at 14:22

Durgesh Tiwari

4,7862426

Let $$I = int frac1+x^2ln xx+x^2ln xdx$$

Divide both Numerator and Denominator by $x^2$

so $$I = int fracfrac1x^2+ln xfrac1x+ln xdx = int fracbigg(frac1x+ln xbigg)-bigg(frac1x-frac1x^2bigg)frac1x+ln xdx$$

so $$I = x-ln bigg|frac1x+ln xbigg|+mathcalC.$$

share|cite|improve this answer

answered Jul 24 at 14:22

Durgesh Tiwari

4,7862426

share|cite|improve this answer

share|cite|improve this answer

answered Jul 24 at 14:22

Durgesh Tiwari

4,7862426

answered Jul 24 at 14:22

Durgesh Tiwari

4,7862426

answered Jul 24 at 14:22

Durgesh Tiwari

4,7862426

4,7862426

add a comment | 

add a comment | 

 

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