Let f(x)=∫x2dx1+x21+1+x2 and f(0)=0. Then the value of f(1) will be

f(x)=∫x2dx1+x21+1+x2 Let x=tan⁡θ ⇒ dx=sec2⁡θdθ∴f(x)=∫x2dx1+x21+1+x2=∫tan2⁡θsec2⁡θdθsec2⁡θ(1+sec⁡θ)=∫tan2⁡θdθ1+sec⁡θ=∫sin2⁡θdθcos⁡θ(1+cos⁡θ)=∫1−cos2⁡θdθcos⁡θ(1+cos⁡θ)=∫(1−cos⁡θ)dθcos⁡θ=∫sec⁡θdθ−∫dθ=log⁡x+1+x2−tan−1⁡x+C Given f(0)=0 or 0=log⁡1−0+C or C=0 or f(1)=log⁡(1+1+1)−tan−1⁡(1)=log⁡(1+2)−π4

Let f(x)=∫x2dx1+x21+1+x2 and f(0)=0. Then the value of f(1) will be

f(x)=∫x2dx1+x21+1+x2 Let x=tan⁡θ ⇒ dx=sec2⁡θdθ∴f(x)=∫x2dx1+x21+1+x2=∫tan2⁡θsec2⁡θdθsec2⁡θ(1+sec⁡θ)=∫tan2⁡θdθ1+sec⁡θ=∫sin2⁡θdθcos⁡θ(1+cos⁡θ)=∫1−cos2⁡θdθcos⁡θ(1+cos⁡θ)=∫(1−cos⁡θ)dθcos⁡θ=∫sec⁡θdθ−∫dθ=log⁡x+1+x2−tan−1⁡x+C Given f(0)=0 or 0=log⁡1−0+C or C=0 or f(1)=log⁡(1+1+1)−tan−1⁡(1)=log⁡(1+2)−π4

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Let $f (x) = \int \frac{x^{2} dx}{(1 + x^{2}) (1 + \sqrt{1 + x^{2}})}$ and $f (0) = 0$ . Then the value of $f (1)$ will be

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\log (1 + \sqrt{2}) - \frac{π}{4}$

$\log (1 + \sqrt{2})$

$\log (1 + \sqrt{2}) + \frac{π}{4}$

$\log (1 + \sqrt{2}) + \frac{π}{2}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$f (x) = \int \frac{x^{2} dx}{(1 + x^{2}) (1 + \sqrt{1 + x^{2}})}$
Let $x = \tan θ ⇒ dx = \sec^{2} θdθ$

$∴ f (x) = \int \frac{x^{2} dx}{(1 + x^{2}) (1 + \sqrt{1 + x^{2}})} = \int \frac{\tan^{2} {θsec}^{2} θdθ}{\sec^{2} θ (1 + \sec θ)} = \int \frac{\tan^{2} θdθ}{1 + \sec θ}$

$\begin{array}{l} = \int \frac{\sin^{2} θdθ}{\cos θ (1 + \cos θ)} = \int \frac{1 - \cos^{2} θdθ}{\cos θ (1 + \cos θ)} = \int \frac{(1 - \cos θ) dθ}{\cos θ} \\ = \int \sec θdθ - \int dθ = \log (x + \sqrt{1 + x^{2}}) - \tan^{- 1} x + C Given f (0) = 0 \\ or 0 = \log 1 - 0 + C or C = 0 or f (1) = \log (1 + \sqrt{1 + 1}) - \tan^{- 1} (1) \\ = \log (1 + \sqrt{2}) - \frac{π}{4} \end{array}$