Integration By Parts

Hello, Ryan!

You integrated ∫t27ln⁡t dt   and got  781t97(9ln⁡t−7)+C\displaystyle \text{You integrated }\int t^{\frac{2}{7}}\ln t\,dt\;\text{ and got }\,\tfrac{7}{81}t^{\frac{9}{7}}(9\ln t - 7) + C

I checked it . . . it’s absolutely correct!\displaystyle \text{I checked it . . . it's absolutely correct!}

. . . . . . Good work!\displaystyle \text{Good work!}
 
Evaluate by I by P ∫1et2/7ln∣t∣dt.\displaystyle Evaluate \ by \ I \ by \ P \ \int_{1}^{e}t^{2/7}ln|t|dt.

Let u = ln∣t∣, then du = dtt.\displaystyle Let \ u \ = \ ln|t|, \ then \ du \ = \ \frac{dt}{t}.

Let dv = t2/7dt, then v = 7t9/79.\displaystyle Let \ dv \ = \ t^{2/7}dt, \ then \ v \ = \ \frac{7t^{9/7}}{9}.

Ergo ∫1et2/7ln∣t∣dt = 79t9/7ln∣t∣]1e−79∫1et2/7dt\displaystyle Ergo \ \int_{1}^{e}t^{2/7}ln|t|dt \ = \ \frac{7}{9}t^{9/7}ln|t|\bigg]_{1}^{e}-\frac{7}{9}\int_{1}^{e} t^{2/7}dt

= 7e9/79−49t9/781]1e = 7e9/79−49e9/781+4981\displaystyle = \ \frac{7e^{9/7}}{9}-\frac{49t^{9/7}}{81}\bigg]_{1}^{e} \ = \ \frac{7e^{9/7}}{9}-\frac{49e^{9/7}}{81}+\frac{49}{81}

= 14e9/7+4981 =˙ 1.23014211103.\displaystyle = \ \frac{14e^{9/7}+49}{81} \ \dot= \ 1.23014211103.
 
soroban said:
Hello, Ryan!

You integrated ∫t27ln⁡t dt   and got  781t97(9ln⁡t−7)+C\displaystyle \text{You integrated }\int t^{\frac{2}{7}}\ln t\,dt\;\text{ and got }\,\tfrac{7}{81}t^{\frac{9}{7}}(9\ln t - 7) + C

I checked it . . . it’s absolutely correct!\displaystyle \text{I checked it . . . it's absolutely correct!}

. . . . . . Good work!\displaystyle \text{Good work!}
Click to expand...



thank you soroban. doesnt get any easier LOL. happy happy joy joy :shock:
 
BigGlenntheHeavy said:
Evaluate by I by P ∫1et2/7ln∣t∣dt.\displaystyle Evaluate \ by \ I \ by \ P \ \int_{1}^{e}t^{2/7}ln|t|dt.

Let u = ln∣t∣, then du = dtt.\displaystyle Let \ u \ = \ ln|t|, \ then \ du \ = \ \frac{dt}{t}.

Let dv = t2/7dt, then v = 7t9/79.\displaystyle Let \ dv \ = \ t^{2/7}dt, \ then \ v \ = \ \frac{7t^{9/7}}{9}.

Ergo ∫1et2/7ln∣t∣dt = 79t9/7ln∣t∣]1e−79∫1et2/7dt\displaystyle Ergo \ \int_{1}^{e}t^{2/7}ln|t|dt \ = \ \frac{7}{9}t^{9/7}ln|t|\bigg]_{1}^{e}-\frac{7}{9}\int_{1}^{e} t^{2/7}dt

= 7e9/79−49t9/781]1e = 7e9/79−49e9/781+4981\displaystyle = \ \frac{7e^{9/7}}{9}-\frac{49t^{9/7}}{81}\bigg]_{1}^{e} \ = \ \frac{7e^{9/7}}{9}-\frac{49e^{9/7}}{81}+\frac{49}{81}

= 14e9/7+4981 =˙ 1.23014211103.\displaystyle = \ \frac{14e^{9/7}+49}{81} \ \dot= \ 1.23014211103.
Click to expand...


its helpful when you dont have a tutor to be right next to you, thats why i like coming here a lot and showing my work. thanx BigGlenntheHeavy. onto the next one. :shock:
 
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