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Vedanshu kumar Grade: 12
```        Solve :

```
7 years ago

SAGAR SINGH - IIT DELHI
879 Points
```										Dear student,
Instead of giving the full solution let me give u a hint:
∫ tan-1x dx
Take u = tan-1x                       dv = x dx
du = [1/(1+ x2)]dx               v = x2/2
∫ tan-1x dx   = uv - ∫ v du
= (x2/2) tan-1 x   - ∫ (x2/2) [1/(1+ x2)]dx
= (x2/2) tan-1 x - (1/2) ∫ (x2 / (1+ x2)]dx
= (x2/2) tan-1 x - (1/2) ∫ (x2 + 1 - 1) / (1+ x2)]dx
= (x2/2) tan-1 x - (1/2) ∫ [1 - [1/ (1+ x2)]] dx
=   (x2/2) tan-1 x - (1/2) [x - tan-1 x] + c

Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.
All the best.
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Sagar Singh
B.Tech, IIT Delhi

```
7 years ago
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